(a) We would like to factor the expression $color{#4257b2}sin^{2}theta-sin theta$. First, we note that the two terms of the expression contain $color{#4257b2}sin theta$, so we can take it as a common factor.

$$
sin^{2}theta-sin theta=sin thetaleft(sin theta-1right)
$$

So the expression $color{#4257b2}sin^{2}theta-sin theta$ can be factored to $boxed{ sin thetaleft(sin theta-1right) }$

(b) We would like to factor the expression $color{#4257b2}cos^{2}theta-2cos theta+1$. First, we note that our expression is a quadratic function on the form $color{#4257b2}a x^{2}+b x+c$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}cos theta$ in our expression, so we can factor it as follows:

$$
begin{align*}
cos^{2}theta-2cos theta+1&=left(cos theta-1right)left(cos theta-1right)
\ \
&=left(cos theta-1right)^{2}
end{align*}
$$

So the expression $color{#4257b2}cos^{2}theta-2cos theta+1$ can be factored to $boxed{ left(cos theta-1right)^{2} }$

(c) We would like to factor the expression $color{#4257b2}3sin^{2}theta-sin theta-2$. First, we note that our expression is a quadratic function on the form $color{#4257b2}a x^{2}+b x+c$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin theta$ in our expression, so we can factor it as follows:

$$
3sin^{2}theta-sin theta-2=left(3sin theta+2right)left(sin theta-1right)
$$

So the expression $color{#4257b2}3sin^{2}theta-sin theta-2$ can be factored to $boxed{ left(3sin theta+2right)left(sin theta-1right) }$

(d) We would like to factor the expression $color{#4257b2}4cos^{2}theta-1$. First, we note that our expression is a quadratic function on the form $color{#4257b2}a^{2} x^{2}-b^{2}$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}cos theta$ in our expression, so we can factor it where $color{#4257b2}a^{2} x^{2} -b^{2}=left(a x-bright)left(a x+bright)$as follows:

$$
4cos^{2}theta-1=left(2cos theta-1right)left(2cos theta+1right)
$$

So the expression $color{#4257b2}4cos^{2}theta-1$ can be factored to $boxed{ left(2cos theta-1right)left(2cos theta+1right) }$

(e) We would like to factor the expression $color{#4257b2}24sin^{2}theta-2sin theta-2$. First, we note that all terms in our expression contain $color{#4257b2}2$, so we can take $color{#4257b2}2$ as a common factor.

$$
24sin^{2}theta-2sin theta-2=2left(12sin^{2}theta-sin theta-1right)
$$

Now we note that our expression is a quadratic function on the form
$color{#4257b2}a x^{2}+b x+c$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin theta$ in our expression, so we can factor it as follows:

$$
begin{align*}
24sin^{2}theta-2sin theta-2&=2left(12sin^{2}theta-sin theta-1right)
\ \
&=2left(4sin theta+1right)left(3sin theta-1right)
end{align*}
$$

So the expression $color{#4257b2}24sin^{2}theta-2sin theta-2$ can be factored to $boxed{ 2left(4sin theta+1right)left(3sin theta-1right) }$

(f) We would like to factor the expression $color{#4257b2}49tan^{2}theta-64$. First, we note that our expression is a quadratic function on the form $color{#4257b2}a^{2} x^{2}-b^{2}$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}tan theta$ in our expression, so we can factor it where $color{#4257b2}a^{2} x^{2} -b^{2}=left(a x-bright)left(a x+bright)$as follows:

$$
49tan^{2}theta-64=left(7tan theta-8right)left(7tan theta+8right)
$$

So the expression $color{#4257b2}49tan^{2}theta-64$ can be factored to $boxed{ left(7tan theta-8right)left(7tan theta+8right) }$

Large{$text{$text{$text{color{#c34632}$(a) sinthetaleft(sin theta-1right)$ $(d) left(2cos theta-1right)left(2cos theta+1right)$
\
\
Large{color{#c34632}$(b) left(cos theta-1right)^{2}$ $(e) 2left(4sin theta+1right)left(3sin theta-1right)$
\
\
Large{color{#c34632}$(c) left(3sintheta+2right)left(sin theta-1right)$ $(f) left(7tan theta-8right)left(7tan theta+8right)$}$}$}$

(a) We would like to solve the equation $color{#4257b2}y^{2}=dfrac{1}{3}$ and then use the same strategy to solve the equation $color{#4257b2}tan^{2}x=dfrac{1}{3}$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that the two equations on the same form where $color{#4257b2}y$ in the first equation is considered to be $color{#4257b2}tan x$ in the second equation, so we will solve the first equation for $color{#4257b2}y$ and then use this strategy to solve the second equation.

For the equation $color{#4257b2}y^{2}=dfrac{1}{3}$

We can take the square root for each side to find the values of $color{#4257b2}y$.

$$
y^{2}=dfrac{1}{3}
$$

$$
y=pm sqrt{dfrac{1}{3}}
$$

$$
y=pm dfrac{1}{sqrt{3}}
$$

So there are two solutions for the first equation $boxed{ y=dfrac{1}{sqrt{3}} } text{or} boxed{ y=-dfrac{1}{sqrt{3}} }$

Now by using the same strategy to solve the equation $color{#4257b2}tan^{2}x=dfrac{1}{3}$ we will find that there are two solutions for $color{#4257b2}tan x$ the first is $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$ and the second is $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$.

Now we have two cases for $color{#4257b2}tan x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}tan =dfrac{1}{sqrt{3}}$

We can take $color{#4257b2}tan^{-1}$ for each side to find the related acute angle for the equation.

$$
tan^{-1}left(tan xright)=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
x=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

Note that $color{#4257b2}tan^{-1}left(tan xright)=x$.

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan dfrac{pi}{6}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions of the equation are exist. We note that $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$ which means that it is is positive, so the solutions in quadrant $1$ and quadrant $3$ where the tangent ratio is positive in these two quadrants and now we can find these solutions using the related acute angle.

$$
x=dfrac{pi}{6} text{or} x=pi+dfrac{pi}{6}
$$

$$
x=dfrac{pi}{6} text{or} x=dfrac{7pi}{6}
$$

So the solutions of the first case are $boxed{ x=dfrac{pi}{6} } text{or} boxed{ x=dfrac{7pi}{6} }$

For $color{#4257b2}tan =-dfrac{1}{sqrt{3}}$

We can take $color{#4257b2}tan^{-1}$ for each side to find the related acute angle for the equation.

$$
tan^{-1}left(tan xright)=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

$$
x=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

Note that $color{#4257b2}tan^{-1}left(tan xright)=x$. Now we will calculate $color{#4257b2}tan^{-1}left(dfrac{1}{sqrt{3}}right)$ to find the acute angle.

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan dfrac{pi}{6}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions of the equation are exist. We note that $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$ which means that it is is negative, so the solutions in quadrant $2$ and quadrant $4$ where the tangent ratio is negative in these two quadrants and now we can find these solutions using the related acute angle.

$$
x=pi-dfrac{pi}{6}=dfrac{5pi}{6} text{or} x=2pi-dfrac{pi}{6}=dfrac{11pi}{6}
$$

So the solutions of the second case are $boxed{ x=dfrac{5pi}{6} } text{or} boxed{ x=dfrac{11pi}{6} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{5pi}{6}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$

(b) We would like to solve the equation $color{#4257b2}y^{2}+y=0$ and then use the same strategy to solve the equation $color{#4257b2}sin^{2}x+sin x=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that the two equations on the same form where $color{#4257b2}y$ in the first equation is considered to be $color{#4257b2}sin x$ in the second equation, so we will solve the first equation for $color{#4257b2}y$ and then use this strategy to solve the second equation.

For the equation $color{#4257b2}y^{2}+y=0$

We can take $color{#4257b2}y$ as a common factor.

$$
y^{2}+y=0
$$

$$
yleft(y+1right)=0
$$

$$
y=0 text{or} y+1=0
$$

$$
y=0 text{or} y=-1
$$

So there are two solutions for the first equation $boxed{ y=0 } text{or} boxed{ y=-1 }$

Now by using the same strategy to solve the equation $color{#4257b2}sin^{2}x+sin x=0$ we will find that there are two solutions for $color{#4257b2}sin x$ the first is $color{#4257b2}sin x=0$ and the second is $color{#4257b2}sin x=-1$.

Now we have two cases for $color{#4257b2}sin x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=0$

We can take $color{#4257b2}sin^{-1}$ for each side to find the related acute angle for the equation.

$$
sin^{-1}left(sin xright)=sin^{-1}left(0right)
$$

$$
x=sin^{-1}left(0right)
$$

$$
x=0, pi text{or} x=2pi
$$

Note that $color{#4257b2}0, pi$ and $color{#4257b2}2pi$ are special angles which we know the values of the sine function of them where $color{#4257b2}sin 0=sin pi=sin 2pi=0$.

So the solutions of the first case are $boxed{ x=0, pi } text{or} boxed{ x=2pi }$

For $color{#4257b2}sin x=-1$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-1right)
$$

$$
x=sin^{-1}left(-1right)
$$

$$
x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{3pi}{2}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{3pi}{2}=-1$.

So the solution of the second case is $boxed{ x=dfrac{3pi}{2} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0, pi, dfrac{3pi}{2}, 2piright}$

(c) We would like to solve the equation $color{#4257b2}y-2yz=0$ and then use the same strategy to solve the equation $color{#4257b2}cos x-2cos xsin x=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that the two equations on the same form where $color{#4257b2}y$ and $color{#4257b2}z$ in the first equation are considered to be $color{#4257b2}cos x$ and $color{#4257b2}sin x$ in the second equation, so we will solve the first equation for $color{#4257b2}y$ and $color{#4257b2}z$ and then use this strategy to solve the second equation.

For the equation $color{#4257b2}y-2yz=0$

We can take $color{#4257b2}y$ as a common factor.

$$
y-2yz=0
$$

$$
yleft(1-2zright)=0
$$

$$
y=0 text{or} 1-2z=0
$$

$$
y=0 text{or} z=dfrac{1}{2}
$$

So there are two solutions for the first equation $boxed{ y=0 } text{or} boxed{ z=dfrac{1}{2} }$

Now by using the same strategy to solve the equation $color{#4257b2}cos x-2cos xsin x=0$ we will find that there are two solutions the first is $color{#4257b2}cos x=0$ and the second is $color{#4257b2}sin x=dfrac{1}{2}$ because we know from the beginning that $color{#4257b2}y$ and $color{#4257b2}z$ in the first equation are considered to be $color{#4257b2}cos x$ and $color{#4257b2}sin x$ in the second equation.

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=0$

We can take $color{#4257b2}cos^{-1}$ for each side.

$$
cos^{-1}left(cos xright)=cos^{-1}left(0right)
$$

$$
x=cos^{-1}left(0right)
$$

$$
x=dfrac{pi}{2} text{or} x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ and $color{#4257b2}dfrac{3pi}{2}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos dfrac{pi}{2}=cos dfrac{3pi}{2}=0$.

So the solutions of the first case are $boxed{ x=dfrac{pi}{2} } text{or} boxed{ x=dfrac{3pi}{2} }$

For $color{#4257b2}sin x=dfrac{1}{2}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{1}{2}right)
$$

$$
x=sin^{-1}left(dfrac{1}{2}right)
$$

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions of the equation are exist. We note that $color{#4257b2}sin x=dfrac{1}{2}$ which means that it is is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these two quadrants and now we can find these solutions using the related acute angle.

$$
x=dfrac{pi}{6} text{or} x=pi-dfrac{pi}{6}
$$

$$
x=dfrac{pi}{6} text{or} x=dfrac{5pi}{6}
$$

So the solutions of the second case are $boxed{ x=dfrac{pi}{6} } text{or} boxed{ x=dfrac{5pi}{6} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{pi}{2}, dfrac{5pi}{6}, dfrac{3pi}{2}right}$

(d) We would like to solve the equation $color{#4257b2}yz=y$ and then use the same strategy to solve the equation $color{#4257b2}tan xsec x=tan x$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that the two equations on the same form where $color{#4257b2}y$ and $color{#4257b2}z$ in the first equation are considered to be $color{#4257b2}tan x$ and $color{#4257b2}sec x$ in the second equation, so we will solve the first equation for $color{#4257b2}y$ and $color{#4257b2}z$ and then use this strategy to solve the second equation.

For the equation $color{#4257b2}yz=y$

$$
yz-y=0
$$

Now we can take $color{#4257b2}y$ as a common factor.

$$
yleft(z-1right)=0
$$

$$
y=0 text{or} z-1=0
$$

$$
y=0 text{or} z=1
$$

So there are two solutions for the first equation $boxed{ y=0 } text{or} boxed{ z=1 }$

Now by using the same strategy to solve the equation $color{#4257b2}tan xsec x=tan x$ we will find that there are two solutions the first is $color{#4257b2}tan x=0$ and the second is $color{#4257b2}sec x=1$ because we know from the beginning that $color{#4257b2}y$ and $color{#4257b2}z$ in the first equation are considered to be $color{#4257b2}tan x$ and $color{#4257b2}sec x$ in the second equation.

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}tan x=0$

$$
tan^{-1}left(tan xright)=tan^{-1}left(0right)
$$

$$
x=tan^{-1}left(0right)
$$

$$
x=0, pi text{or} x=2pi
$$

Note that $color{#4257b2}0, pi$ and $color{#4257b2}2pi$ are special angles which we know the values of the tangent function of them where $color{#4257b2}tan 0=tan pi=tan 2pi=0$.

So the solutions of the first case are $boxed{ x=0, pi } text{or} boxed{ x=2pi }$

For $color{#4257b2}sec x=1$

Since we know that $color{#4257b2}cos x=dfrac{1}{sec x}$, so we can use this identity in our equation as follows:

$$
sec x=1
$$

$$
dfrac{1}{sec x}=dfrac{1}{1}
$$

$$
cos x=1
$$

Now we can take $color{#4257b2}cos^{-1}$ for each case to find the values of $color{#4257b2}x$

$$
cos^{-1}left(cos xright)=cos^{-1}left(1right)
$$

$$
x=cos^{-1}left(1right)
$$

$$
x=0 text{or} x=2pi
$$

Note that $color{#4257b2}0$ and $color{#4257b2}2pi$ are special angles which we know the value of the cosine function of it where $color{#4257b2}cos 0=cos 2pi=1$.

So the solutions of the second case are $boxed{ x=0 } text{or} boxed{ x=2pi }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0, pi, 2piright}$

Large{$text{$text{$text{$text{color{#c34632}$(a) y=dfrac{1}{sqrt{3}} {color{Black}text{or}} y=-dfrac{1}{sqrt{3}}$, $x=left{dfrac{pi}{6}, dfrac{5pi}{6}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$
\
\
\
Large{color{#c34632}$(b) y=0 {color{Black}text{or}} y=-1$, $x=left{0, pi, dfrac{3pi}{2}, 2piright}$
\
\
\
Large{color{#c34632}$(c) y=0 {color{Black}text{or}} z=dfrac{1}{2}$, $x=left{dfrac{pi}{6}, dfrac{pi}{2}, dfrac{5pi}{6}, dfrac{3pi}{2}right}$
\
\
\
Large{color{#c34632}$(d) y=0 {color{Black}text{or}} y=1$, $x=left{0, pi, 2piright}$}$}$}$}$

(a) We would like to solve the equation $color{#4257b2}6y^{2}-y-1=0$. First, we note that our equation is a quadratic equation on the form $color{#4257b2}a y^{2}+b y+c=0$, so we can factor to find the values of $color{#4257b2}y$.

$$
6y^{2}-y-1=0
$$

$$
left(2y-1right)left(3y+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}y$.

$$
2y-1=0 text{or} 3y+1=0
$$

$$
2y=1 text{or} 3y=-1
$$

$$
y=dfrac{1}{2} text{or} y=-dfrac{1}{3}
$$

So there are two solutions for the equation $boxed{ y=dfrac{1}{2} } text{or} boxed{ y=-dfrac{1}{3} }$

(b) We would like to solve the equation $color{#4257b2}6cos^{2}x-cos x-1=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that our equation is a quadratic equation on the same form for the equation in question (a) where $color{#4257b2}y$ in the first equation is considered to be $color{#4257b2}cos x$ in this equation, so we can use the same strategy to solve this equation.

$$
6cos^{2}x-cos x-1=0
$$

$$
left(2cos x-1right)left(3cos x+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}y$.

$$
2cos x-1=0 text{or} 3cos x+1=0
$$

$$
2cos x=1 text{or} 3cos x=-1
$$

$$
cos x=dfrac{1}{2} text{or} cos x=-dfrac{1}{3}
$$

Now we have two cases for $color{#4257b2}cos x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=dfrac{1}{2}$

Now we can take $color{#4257b2}cos^{-1}$ fro each side to find the related acute angle.

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=cos^{-1}left(dfrac{1}{2}right)
$$

Note that $color{#4257b2}cos^{-1}left(cos xright)=x$.

$$
x=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions of the equation are exist. We note that $color{#4257b2}cos x=dfrac{1}{2}$ which means that it is is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these two quadrants and now we can find these solutions using the related acute angle.

$$
x=dfrac{pi}{3} text{or} x=2pi-dfrac{pi}{3}
$$

$$
x=dfrac{pi}{3} text{or} x=dfrac{5pi}{3}
$$

So the solutions of the first case are $boxed{ x=dfrac{pi}{3} } text{or} boxed{ x=dfrac{5pi}{3} }$

For $color{#4257b2}cos x=-dfrac{1}{3}$

Now we can take $color{#4257b2}cos^{-1}$ fro each side to find the related acute angle.

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{1}{3}right)
$$

$$
x=cos^{-1}left(-dfrac{1}{3}right)
$$

Now we will use the calculator to determine $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ to find the related acute angle.

$$
x=1.23 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{1}{3}$, so the next step is to know in which quadrants the solutions of the equation are exist. We note that $color{#4257b2}cos x=-dfrac{1}{3}$ which means that it is is negative, so the solutions in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these two quadrants and now we can find these solutions using the related acute angle.

$$
x=pi-1.23 text{or} x=pi+1.23
$$

$$
x=1.91 text{or} x=4.37
$$

So the solutions of the second case are $boxed{ x=1.91 text{radians} } text{or} boxed{ x=4.37 text{radians} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{1.91, 4.37, dfrac{pi}{3}, dfrac{5pi}{3}right}$

$$text{$text{$text{color{#c34632}(a) y=dfrac{1}{2} {color{Black}text{or}} y=-dfrac{1}{3}\
\
\
Large{color{BrickRed$
$(b) x=left{1.91, 4.37, dfrac{pi}{3}, dfrac{5pi}{3}right}$}$}$}$

(a) We would like to solve the equation $color{#4257b2}sin^{2}theta=1$ for $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can take the square root for each side to find the values of $color{#4257b2}sin theta$.

$$
sin^{2}theta=1
$$

$$
sin theta=pm sqrt{1}
$$

$$
sin theta=pm 1
$$

Now we have two cases for $color{#4257b2}sin theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}sin theta=1$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(1right)
$$

$$
theta=sin^{-1}left(1right)
$$

$$
theta=90text{textdegree}
$$

Note that $color{#4257b2}90text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 90text{textdegree}=1$.

So the solution of the first case is $boxed{ theta=90text{textdegree} }$

For $color{#4257b2}sin theta=-1$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(-1right)
$$

$$
theta=sin^{-1}left(-1right)
$$

$$
theta=270text{textdegree}
$$

Note that $color{#4257b2}270text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 270text{textdegree}=-1$.

So the solution of the second case is $boxed{ theta=270text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{90text{textdegree}, 270text{textdegree}right}$

(b) We would like to solve the equation $color{#4257b2}cos^{2}theta=1$ for $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can take the square root for each side to find the values of $color{#4257b2}cos theta$.

$$
cos^{2}theta=1
$$

$$
cos theta=pm sqrt{1}
$$

$$
cos theta=pm 1
$$

Now we have two cases for $color{#4257b2}cos theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}cos theta=1$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(1right)
$$

$$
theta=cos^{-1}left(1right)
$$

$$
theta=0text{textdegree} text{or} theta=360text{textdegree}
$$

Note that $color{#4257b2}0text{textdegree}$ and $color{#4257b2}360text{textdegree}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 0text{textdegree}=cos 360text{textdegree}=1$.

So the solutions of the first case are $boxed{ theta=0text{textdegree} } text{or} boxed{ theta=360text{textdegree} }$

For $color{#4257b2}cos theta=-1$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(-1right)
$$

$$
theta=cos^{-1}left(-1right)
$$

$$
theta=180text{textdegree}
$$

Note that $color{#4257b2}180text{textdegree}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos 180text{textdegree}=-1$.

So the solution of the second case is $boxed{ theta=180text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{0text{textdegree}, 180text{textdegree}, 360text{textdegree}right}$

(c) We would like to solve the equation $color{#4257b2}tan^{2}theta=1$ for $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can take the square root for each side to find the values of $color{#4257b2}tan theta$.

$$
tan^{2}theta=1
$$

$$
tan theta=pm sqrt{1}=pm 1
$$

Now we have two cases for $color{#4257b2}tan theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}tan theta=1$

$$
tan^{-1}left(tan thetaright)=tan^{-1}left(1right)
$$

$$
theta=tan^{-1}left(1right)
$$

$$
theta=45text{textdegree}
$$

Note that $color{#4257b2}45text{textdegree}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan 45text{textdegree}=1$.

Now we found the related acute angle for the equation $color{#4257b2}tan theta=1$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan theta=1$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $3$ and now we can use the related acute angle to find the solutions.

$$
theta=45text{textdegree} text{or} theta=180text{textdegree}+45text{textdegree}
$$

$$
theta=45text{textdegree} text{or} theta=225text{textdegree}
$$

So the solutions of the first case are $boxed{ theta=45text{textdegree} } text{or} boxed{ theta=225text{textdegree} }$

For $color{#4257b2}tan theta=-1$

$$
tan^{-1}left(tan thetaright)=tan^{-1}left(-1right)
$$

$$
theta=tan^{-1}left(-1right)
$$

Now we will calculate $color{#4257b2}tan^{-1}(1)$ to find the related acute angle.

$$
theta=45text{textdegree}
$$

Note that $color{#4257b2}45text{textdegree}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan 45text{textdegree}=1$.

Now we found the related acute angle for the equation $color{#4257b2}tan theta=-1$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan theta=-1$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
theta=180text{textdegree}-45text{textdegree} text{or} theta=360text{textdegree}-45text{textdegree}
$$

$$
theta=135text{textdegree} text{or} theta=315text{textdegree}
$$

So the solutions of the second case are $boxed{ theta=135text{textdegree} } text{or} boxed{ theta=315text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{45text{textdegree}, 135text{textdegree}, 225text{textdegree}, 315text{textdegree}right}$

(d) We would like to solve the equation $color{#4257b2}4cos^{2}theta=1$ for $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can divide the two sides by $color{#4257b2}4$ to make $color{#4257b2}cos theta$ in the left side alone.

$$
4cos^{2}theta=1
$$

$$
cos^{2}theta=dfrac{1}{4}
$$

Now we can take the square root for each side to find the values of $color{#4257b2}cos theta$.

$$
cos theta=pm sqrt{dfrac{1}{4}}
$$

$$
cos theta=pm dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}cos theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}cos theta=dfrac{1}{2}$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(dfrac{1}{2}right)
$$

$$
theta=cos^{-1}left(dfrac{1}{2}right)
$$

$$
theta=60text{textdegree}
$$

Note that $color{#4257b2}60text{textdegree}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos 60text{textdegree}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos theta=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos theta=dfrac{1}{2}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
theta=60text{textdegree} text{or} theta=360text{textdegree}-60text{textdegree}
$$

$$
theta=60text{textdegree} text{or} theta=300text{textdegree}
$$

So the solutions of the first case are $boxed{ theta=60text{textdegree} } text{or} boxed{ theta=300text{textdegree} }$

For $color{#4257b2}cos theta=-dfrac{1}{2}$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(-dfrac{1}{2}right)
$$

$$
theta=cos^{-1}left(-dfrac{1}{2}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ to find the related acute angle.

$$
theta=60text{textdegree}
$$

Note that $color{#4257b2}60text{textdegree}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos 60text{textdegree}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos theta=-dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos theta=-dfrac{1}{2}$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $3$ and now we can use the related acute angle to find the solutions.

$$
theta=180text{textdegree}-60text{textdegree} text{or} theta=180text{textdegree}+60text{textdegree}
$$

$$
theta=120text{textdegree} text{or} theta=240text{textdegree}
$$

So the solutions of the second case are $boxed{ theta=120text{textdegree} } text{or} boxed{ theta=240text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{60text{textdegree}, 120text{textdegree}, 240text{textdegree}, 300text{textdegree}right}$

(e) We would like to solve the equation $color{#4257b2}3tan^{2}theta=1$ for $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can divide the two sides by $color{#4257b2}3$ to make $color{#4257b2}tan theta$ in the left side alone.

$$
3tan^{2}theta=1
$$

$$
tan^{2}theta=dfrac{1}{3}
$$

Now we can take the square root for each side to find the values of $color{#4257b2}tan theta$.

$$
tan theta=pm sqrt{dfrac{1}{3}}
$$

$$
tan theta=pm dfrac{1}{sqrt{3}}
$$

Now we have two cases for $color{#4257b2}tan theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}tan theta=dfrac{1}{sqrt{3}}$

$$
tan^{-1}left(tan thetaright)=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
theta=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
theta=30text{textdegree}
$$

Note that $color{#4257b2}30text{textdegree}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan 30text{textdegree}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan theta=dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan theta=dfrac{1}{sqrt{3}}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $3$ and now we can use the related acute angle to find the solutions.

$$
theta=30text{textdegree} text{or} theta=180text{textdegree}+30text{textdegree}
$$

$$
theta=30text{textdegree} text{or} theta=210text{textdegree}
$$

So the solutions of the first case are $boxed{ theta=30text{textdegree} } text{or} boxed{ theta=210text{textdegree} }$

For $color{#4257b2}tan theta=-dfrac{1}{sqrt{3}}$

$$
tan^{-1}left(tan thetaright)=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

$$
theta=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

Now we will calculate $color{#4257b2}tan^{-1}left(dfrac{1}{sqrt{3}}right)$ to find the related acute angle.

$$
theta=30text{textdegree}
$$

Note that $color{#4257b2}30text{textdegree}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan 30text{textdegree}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan theta=-dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan theta=-dfrac{1}{sqrt{3}}$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
theta=180text{textdegree}-30text{textdegree} text{or} theta=360text{textdegree}-30text{textdegree}
$$

$$
theta=150text{textdegree} text{or} theta=330text{textdegree}
$$

So the solutions of the second case are $boxed{ theta=150text{textdegree} } text{or} boxed{ theta=330text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{30text{textdegree}, 150text{textdegree}, 210text{textdegree}, 330text{textdegree}right}$

(f) We would like to solve the equation $color{#4257b2}2sin^{2}theta=1$ for $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can divide the two sides by $color{#4257b2}2$ to make $color{#4257b2}sin theta$ in the left side alone.

$$
2sin^{2}theta=1
$$

$$
sin^{2}theta=dfrac{1}{2}
$$

Now we can take the square root for each side to find the values of $color{#4257b2}sin theta$.

$$
sin theta=pm sqrt{dfrac{1}{2}}
$$

$$
sin theta=pm dfrac{1}{sqrt{2}}
$$

Now we have two cases for $color{#4257b2}sin theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}sin theta=dfrac{1}{sqrt{2}}$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
theta=sin^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
theta=45text{textdegree}
$$

Note that $color{#4257b2}45text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 45text{textdegree}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}sin theta=dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin theta=dfrac{1}{sqrt{2}}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $2$ and now we can use the related acute angle to find the solutions.

$$
theta=45text{textdegree} text{or} theta=180text{textdegree}-45text{textdegree}
$$

$$
theta=45text{textdegree} text{or} theta=135text{textdegree}
$$

So the solutions of the first case are $boxed{ theta=45text{textdegree} } text{or} boxed{ theta=135text{textdegree} }$

For $color{#4257b2}sin theta=-dfrac{1}{sqrt{2}}$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

$$
theta=sin^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{sqrt{2}}right)$ to find the related acute angle.

$$
theta=45text{textdegree}
$$

Note that $color{#4257b2}45text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 45text{textdegree}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}sin theta=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin theta=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions are in quadrant $3$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
theta=180text{textdegree}+45text{textdegree} text{or} theta=360text{textdegree}-45text{textdegree}
$$

$$
theta=225text{textdegree} text{or} theta=315text{textdegree}
$$

So the solutions of the second case are $boxed{ theta=225text{textdegree} } text{or} boxed{ theta=315text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{45text{textdegree}, 135text{textdegree}, 225text{textdegree}, 315text{textdegree}right}$

Large{$text{$text{$text{color{#c34632}$(a) theta=left{90text{textdegree}, 270text{textdegree}right}$ $(d) theta=left{60text{textdegree}, 120text{textdegree}, 240text{textdegree}, 300text{textdegree}right}$
\
\
\
Large{color{#c34632}$(b) theta=left{0text{textdegree}, 180text{textdegree}, 360text{textdegree}right}$ $(e) theta=left{30text{textdegree}, 150text{textdegree}, 210text{textdegree}, 330text{textdegree}right}$
\
\
\
Large{color{#c34632}$(c) theta=left{45text{textdegree}, 135text{textdegree}, 225text{textdegree}, 315text{textdegree}right}$ $(f) theta=left{45text{textdegree}, 135text{textdegree}, 225text{textdegree}, 315text{textdegree}right}$}$}$}$

(a) We would like to solve the equation $color{#4257b2}sin xcos x=0$ for $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
sin xcos x=0
$$

$$
sin x=0 text{or} cos x=0
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=0$

$$
sin^{-1}left(sin xright)=sin^{-1}left(0right)
$$

$$
x=sin^{-1}left(0right)
$$

$$
x=0text{textdegree}, 180text{textdegree} text{or} x=360text{textdegree}
$$

Note that $color{#4257b2}0text{textdegree}, 180text{textdegree}$ and $color{#4257b2}360text{textdegree}$ are special angles which we know the values of the sine function of them where $color{#4257b2}sin 0text{textdegree}=sin 180text{textdegree}=sin 360text{textdegree}=0$.

So the solutions of the first case are $boxed{ x=0text{textdegree}, 180text{textdegree} } text{or} boxed{ x=360text{textdegree} }$

For $color{#4257b2}cos x=0$

$$
cos^{-1}left(cos xright)=cos^{-1}left(0right)
$$

$$
x=cos^{-1}left(0right)
$$

$$
x=90text{textdegree} text{or} x=270text{textdegree}
$$

Note that $color{#4257b2}90text{textdegree}$ and $color{#4257b2}270text{textdegree}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 90text{textdegree}=cos 270text{textdegree}=0$.

So the solutions of the second case are $boxed{ x=90text{textdegree} } text{or} boxed{ x=270text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0text{textdegree}, 90text{textdegree}, 180text{textdegree}, 270text{textdegree}, 360text{textdegree}right}$

(b) We would like to solve the equation $color{#4257b2}sin xleft(cos x-1right)=0$
for $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
sin xleft(cos x-1right)=0
$$

$$
sin x=0 text{or} cos x-1=0
$$

$$
sin x=0 text{or} cos x=1
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=0$

$$
sin^{-1}left(sin xright)=sin^{-1}left(0right)
$$

$$
x=sin^{-1}left(0right)
$$

$$
x=0text{textdegree}, 180text{textdegree} text{or} x=360text{textdegree}
$$

Note that $color{#4257b2}0text{textdegree}, 180text{textdegree}$ and $color{#4257b2}360text{textdegree}$ are special angles which we know the values of the sine function of them where $color{#4257b2}sin 0text{textdegree}=sin 180text{textdegree}=sin 360text{textdegree}=0$.

So the solutions of the first case are $boxed{ x=0text{textdegree}, 180text{textdegree} } text{or} boxed{ x=360text{textdegree} }$

For $color{#4257b2}cos x=1$

$$
cos^{-1}left(cos xright)=cos^{-1}left(1right)
$$

$$
x=cos^{-1}left(1right)
$$

$$
x=0text{textdegree} text{or} x=360text{textdegree}
$$

Note that $color{#4257b2}0text{textdegree}$ and $color{#4257b2}360text{textdegree}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 0text{textdegree}=cos 360text{textdegree}=1$.

So the solutions of the second case are $boxed{ x=0text{textdegree} } text{or} boxed{ x=360text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0text{textdegree}, 180text{textdegree}, 360text{textdegree}right}$

(c) We would like to solve the equation $color{#4257b2}left(sin x+1right)cos x=0$ for
$color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(sin x+1right)cos x=0
$$

$$
sin x+1=0 text{or} cos x=0
$$

$$
sin x=-1 text{or} cos x=0
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=-1$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-1right)
$$

$$
x=sin^{-1}left(-1right)
$$

$$
x=270text{textdegree}
$$

Note that $color{#4257b2}270text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 270text{textdegree}=-1$.

So the solution of the first case is $boxed{ x=270text{textdegree} }$

For $color{#4257b2}cos x=0$

$$
cos^{-1}left(cos xright)=cos^{-1}left(0right)
$$

$$
x=cos^{-1}left(0right)
$$

$$
x=90text{textdegree} text{or} x=270text{textdegree}
$$

Note that $color{#4257b2}90text{textdegree}$ and $color{#4257b2}270text{textdegree}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 90text{textdegree}=cos 270text{textdegree}=0$.

So the solutions of the second case are $boxed{ x=90text{textdegree} } text{or} boxed{ x=270text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{90text{textdegree}, 270text{textdegree}right}$

(d) We would like to solve the equation $color{#4257b2}cos xleft(2sin x-sqrt{3}right)=0$
for $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
cos xleft(2sin x-sqrt{3}right)=0
$$

$$
cos x=0 text{or} 2sin x-sqrt{3}=0
$$

$$
cos x=0 text{or} sin x=dfrac{sqrt{3}}{2}
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=0$

$$
cos^{-1}left(cos xright)=cos^{-1}left(0right)
$$

$$
x=cos^{-1}left(0right)
$$

$$
x=90text{textdegree} text{or} x=270text{textdegree}
$$

Note that $color{#4257b2}90text{textdegree}$ and $color{#4257b2}270text{textdegree}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 90text{textdegree}=cos 270text{textdegree}=0$.

So the solutions of the first case are $boxed{ x=90text{textdegree} } text{or} boxed{ x=270text{textdegree} }$

For $color{#4257b2}sin x=dfrac{sqrt{3}}{2}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{sqrt{3}}{2}right)
$$

$$
x=sin^{-1}left(dfrac{sqrt{3}}{2}right)
$$

$$
x=60text{textdegree}
$$

Note that $color{#4257b2}60text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 60text{textdegree}=dfrac{sqrt{3}}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=dfrac{sqrt{3}}{2}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $2$ and now we can use the related acute angle to find the solutions.

$$
x=60text{textdegree} text{or} x=180text{textdegree}-60text{textdegree}
$$

$$
x=60text{textdegree} text{or} x=120text{textdegree}
$$

So the solutions of the second case are $boxed{ x=60text{textdegree} } text{or} boxed{ x=120text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{60text{textdegree}, 90text{textdegree}, 120text{textdegree}, 270text{textdegree}right}$

(e) We would like to solve the equation $color{#4257b2}left(sqrt{2}sin x-1right)left(sqrt{2}sin x+1right)=0$
for $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(sqrt{2}sin x-1right)left(sqrt{2}sin x+1right)=0
$$

$$
sqrt{2} sin x-1=0 text{or} sqrt{2} sin x+1=0
$$

$$
sin x=dfrac{1}{sqrt{2}} text{or} sin x=-dfrac{1}{sqrt{2}}
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=dfrac{1}{sqrt{2}}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
x=sin^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
x=45text{textdegree}
$$

Note that $color{#4257b2}45text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 45text{textdegree}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=dfrac{1}{sqrt{2}}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $2$ and now we can use the related acute angle to find the solutions.

$$
x=45text{textdegree} text{or} x=180text{textdegree}-45text{textdegree}
$$

$$
x=45text{textdegree} text{or} x=135text{textdegree}
$$

So the solutions of the first case are $boxed{ x=45text{textdegree} } text{or} boxed{ x=135text{textdegree} }$

For $color{#4257b2}sin x=-dfrac{1}{sqrt{2}}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

$$
x=sin^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

Now we can calculate $color{#4257b2}sin^{-1}left(dfrac{1}{sqrt{2}}right)$ to find the related acute angle.

$$
x=45text{textdegree}
$$

Note that $color{#4257b2}45text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 45text{textdegree}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions are in quadrant $3$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
x=180text{textdegree}+45text{textdegree} text{or} x=360text{textdegree}-45text{textdegree}
$$

$$
x=225text{textdegree} text{or} x=315text{textdegree}
$$

So the solutions of the second case are $boxed{ x=225text{textdegree} } text{or} boxed{ x=315text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{45text{textdegree}, 135text{textdegree}, 225text{textdegree}, 315text{textdegree}right}$

(f) We would like to solve the equation $color{#4257b2}left(sin x-1right)left(cos x+1right)=0$
for $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(sin x-1right)left(cos x+1right)=0
$$

$$
sin x-1=0 text{or} cos x+1=0
$$

$$
sin x=1 text{or} cos x=-1
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=1$

$$
sin^{-1}left(sin xright)=sin^{-1}left(1right)
$$

$$
x=sin^{-1}left(1right)
$$

$$
x=90text{textdegree}
$$

Note that $color{#4257b2}90text{textdegree}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 90text{textdegree}=1$.

So the solution of the first case is $boxed{ x=90text{textdegree} }$

For $color{#4257b2}cos x=-1$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-1right)
$$

$$
x=cos^{-1}left(-1right)
$$

$$
x=180text{textdegree}
$$

Note that $color{#4257b2}180text{textdegree}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos 180text{textdegree}=-1$.

So the solution of the second case is $boxed{ x=180text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{90text{textdegree}, 180text{textdegree}right}$

Large{$text{$text{$text{color{#c34632}$(a) x=left{0text{textdegree}, 90text{textdegree}, 180text{textdegree}, 270text{textdegree}, 360text{textdegree}right}$ $(d) x=left{60text{textdegree}, 90text{textdegree}, 120text{textdegree}, 270text{textdegree}right}$
\
\
\
Large{color{#c34632}$(b) x=left{0text{textdegree}, 180text{textdegree}, 360text{textdegree}right}$ $(e) x=left{45text{textdegree}, 135text{textdegree}, 225text{textdegree}, 315text{textdegree}right}$
\
\
\
Large{color{#c34632}$(c) x=left{90text{textdegree}, 270text{textdegree}right}$ $(f) theta=left{90text{textdegree}, 180text{textdegree}right}$}$}$}$

(a) We would like to solve the equation $color{#4257b2}left(2sin x-1right)cos x=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(2sin x-1right)cos x=0
$$

$$
2sin x-1=0 text{or} cos x=0
$$

$$
sin x=dfrac{1}{2} text{or} cos x=0
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=0$

$$
cos^{-1}left(cos xright)=cos^{-1}left(0right)
$$

$$
x=cos^{-1}left(0right)
$$

$$
x=dfrac{pi}{2} text{or} x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ and $color{#4257b2}dfrac{3pi}{2}$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos dfrac{pi}{2}=cos dfrac{3pi}{2}=0$.

So the solutions of the first case are $boxed{ x=dfrac{pi}{2} } text{or} boxed{ x=dfrac{3pi}{2} }$

For $color{#4257b2}sin x=dfrac{1}{2}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{1}{2}right)
$$

$$
x=sin^{-1}left(dfrac{1}{2}right)
$$

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{6} text{or} x=pi-dfrac{pi}{6}
$$

$$
x=dfrac{pi}{6} text{or} x=dfrac{5pi}{6}
$$

So the solutions of the second case are $boxed{ x=dfrac{pi}{6} } text{or} boxed{ x=dfrac{5pi}{6} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{pi}{2}, dfrac{5pi}{6}, dfrac{3pi}{2}right}$

(b) We would like to solve the equation $color{#4257b2}left(sin x+1right)^{2}$ for $color{#4257b2}0 leq x leq 2pi$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(sin x+1right)^{2}=0
$$

$$
sin x+1=0 text{or} sin x+1=0
$$

$$
sin x=-1 text{or} sin x=-1
$$

Now we have two cases but the two cases are the same, so we can solve one of them to find the values of $color{#4257b2}x$.

$$
sin x=-1
$$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-1right)
$$

$$
x=sin^{-1}left(-1right)
$$

$$
x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{3pi}{2}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{3pi}{2}=-1$.

So the solution of the equation is $color{#4257b2}left{x=dfrac{3pi}{2}right}$

(c) We would like to solve the equation $color{#4257b2}left(2cos x+sqrt{3}right)sin x=0$
for $color{#4257b2}0 leq x leq 2pi$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(2cos x+sqrt{3}right)sin x=0
$$

$$
2cos x+sqrt{3}=0 text{or} sin x=0
$$

$$
cos x=-dfrac{sqrt{3}}{2} text{or} sin x=0
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=0$

$$
sin^{-1}left(sin xright)=sin^{-1}left(0right)
$$

$$
x=sin^{-1}left(0right)
$$

$$
x=0, pi text{or} x=2pi
$$

Note that $color{#4257b2}0, pi$ and $color{#4257b2}2pi$ are special angles which we know the values of the sine function of them where $color{#4257b2}sin 0=sin pi=sin 2pi=0$.

So the solutions of the first case are $boxed{ x=0, pi } text{or} boxed{ x=2pi }$

For $color{#4257b2}cos x=-dfrac{sqrt{3}}{2}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{sqrt{3}}{2}right)
$$

$$
x=cos^{-1}left(-dfrac{sqrt{3}}{2}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{sqrt{3}}{2}right)$ to find the related acute angle.

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{6}=dfrac{sqrt{3}}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{sqrt{3}}{2}$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{6} text{or} x=pi+dfrac{pi}{6}
$$

$$
x=dfrac{5pi}{6} text{or} x=dfrac{7pi}{6}
$$

So the solutions of the second case are $boxed{ x=dfrac{5pi}{6} } text{or} boxed{ x=dfrac{7pi}{6} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0, dfrac{5pi}{6}, pi, dfrac{7pi}{6}, 2piright}$

(d) We would like to solve the equation $color{#4257b2}left(2cos x-1right)left(2sin x+sqrt{3}right)=0$
for $color{#4257b2}0 leq x leq 2pi$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(2cos x-1right)left(2sin x+sqrt{3}right)=0
$$

$$
2cos x-1=0 text{or} 2sin x+sqrt{3}=0
$$

$$
cos x=dfrac{1}{2} text{or} sin x=-dfrac{sqrt{3}}{2}
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=dfrac{1}{2}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{3} text{or} x=2pi-dfrac{pi}{3}
$$

$$
x=dfrac{pi}{3} text{or} x=dfrac{5pi}{3}
$$

So the solutions of the first case are $boxed{ x=dfrac{pi}{3} } text{or} boxed{ x=dfrac{5pi}{3} }$

For $color{#4257b2}sin x=-dfrac{sqrt{3}}{2}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-dfrac{sqrt{3}}{2}right)
$$

$$
x=sin^{-1}left(-dfrac{sqrt{3}}{2}right)
$$

Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{sqrt{3}}{2}right)$ to find the related acute angle.

$$
x=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{sqrt{3}}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=-dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=-dfrac{sqrt{3}}{2}$ which means that it is negative, so the solutions in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi+dfrac{pi}{3} text{or} x=2pi-dfrac{pi}{3}
$$

$$
x=dfrac{4pi}{3} text{or} x=dfrac{5pi}{3}
$$

So the solutions of the second case are $boxed{ x=dfrac{4pi}{3} } text{or} boxed{ x=dfrac{5pi}{3} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{3}, dfrac{4pi}{3}, dfrac{5pi}{3}right}$

(e) We would like to solve the equation $color{#4257b2}left(sqrt{2} cos x-1right)left(sqrt{2} cos x+1right)=0$
for $color{#4257b2}0 leq x leq 2pi$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(sqrt{2} cos x-1right)left(sqrt{2} cos x+1right)=0
$$

$$
sqrt{2} cos x-1=0 text{or} sqrt{2} cos x+1=0
$$

$$
cos x=dfrac{1}{sqrt{2}} text{or} cos x=-dfrac{1}{sqrt{2}}
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=dfrac{1}{sqrt{2}}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
x=cos^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
x=dfrac{pi}{4}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{4}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{1}{sqrt{2}}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{4} text{or} x=2pi-dfrac{pi}{4}
$$

$$
x=dfrac{pi}{4} text{or} x=dfrac{7pi}{4}
$$

So the solutions of the first case are $boxed{ x=dfrac{pi}{4} } text{or} boxed{ x=dfrac{7pi}{4} }$

For $color{#4257b2}cos x=-dfrac{1}{sqrt{2}}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

$$
x=cos^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{sqrt{2}}right)$ to find the related acute angle.

$$
x=dfrac{pi}{4}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{4}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $3$ and now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{4} text{or} x=pi+dfrac{pi}{4}
$$

$$
x=dfrac{3pi}{4} text{or} x=dfrac{5pi}{4}
$$

So the solutions of the first case are $boxed{ x=dfrac{3pi}{4} } text{or} boxed{ x=dfrac{5pi}{4} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{4}, dfrac{3pi}{4}, dfrac{5pi}{4}, dfrac{7pi}{4}right}$

(f) We would like to solve the equation $color{#4257b2}left(sin x+1right)left(cos x-1right)=0$
for $color{#4257b2}0 leq x leq 2pi$. First, we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
left(sin x+1right)left(cos x-1right)=0
$$

$$
sin x+1=0 text{or} cos x-1=0
$$

$$
sin x=-1 text{or} cos x=1
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=-1$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-1right)
$$

$$
x=sin^{-1}left(-1right)
$$

$$
x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{3pi}{2}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{3pi}{2}=-1$.

So the solution of the first case is $boxed{ x=dfrac{3pi}{2} }$

For $color{#4257b2}cos x=1$

$$
cos^{-1}left(cos xright)=cos^{-1}left(1right)
$$

$$
x=cos^{-1}left(1right)
$$

$$
x=0 text{or} x=2pi
$$

Note that $color{#4257b2}0$ and $color{#4257b2}2pi$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 0=cos 2pi=1$.

So the solution of the second case is $boxed{ x=0 } text{or} boxed{ x=2pi }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0, dfrac{3pi}{2}. 2piright}$

$$
text{color{#c34632}$(a) x=left{dfrac{pi}{6}, dfrac{pi}{2}, dfrac{5pi}{6}, dfrac{3pi}{2}right}$ $(d) x=left{dfrac{pi}{3}, dfrac{4pi}{3}, dfrac{5pi}{3}right}$
\
\
\
color{#c34632}$(b) x=left{dfrac{3pi}{2}right}$ $(e) x=left{dfrac{pi}{4}, dfrac{3pi}{4}, dfrac{5pi}{4}, dfrac{7pi}{4}right}$
\
\
\
color{#c34632}$(c) x=left{0, dfrac{5pi}{6}, pi, dfrac{7pi}{6}, 2piright}$ $(f) x=left{0, dfrac{3pi}{2}. 2piright}$}
$$

(a) We would like to solve the equation $color{#4257b2}2cos^{2}theta+cos theta-1=0$ to the nearest hundredth where $color{#4257b2}0 leq theta leq 2pi$. First, we note that our equation is a quadratic equation on the form $color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}cos theta$ in our equation, so we can factor to find the values of $color{#4257b2}cos theta$.

$$
2cos^{2}theta+cos theta-1=0
$$

$$
left(cos theta+1right)left(2cos theta-1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos theta$.

$$
cos theta+1=0 text{or} 2cos theta-1=0
$$

$$
cos theta=-1 text{or} cos theta=dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}cos theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}cos theta=-1$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(-1right)
$$

$$
theta=cos^{-1}left(-1right)
$$

$$
theta=pi=3.14 text{radians}
$$

Note that $color{#4257b2}pi$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos pi=-1$.

So the solution of the first case is $boxed{ theta=3.14 text{radians} }$

For $color{#4257b2}cos theta=dfrac{1}{2}$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(dfrac{1}{2}right)
$$

$$
theta=cos^{-1}left(dfrac{1}{2}right)
$$

$$
theta=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos theta=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos theta=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=dfrac{pi}{3} text{or} theta=2pi-dfrac{pi}{3}
$$

$$
theta=dfrac{pi}{3}=1.05 text{radians} text{or} theta=dfrac{5pi}{3}=5.24 text{radians}
$$

So the solutions of the second case are $boxed{ theta=1.05 text{radians} } text{or} boxed{ theta=5.24 text{radians} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{1.05 text{radians}, 3.14 text{radians}, 5.24 text{radians}right}$

(b) We would like to solve the equation $color{#4257b2}2sin^{2}theta=1-sin theta$ to the nearest hundredth where $color{#4257b2}0 leq theta leq 2pi$. First, we can add $color{#4257b2}sin theta-1$ to each side to make the right side equals zero.

$$
2sin^{2}theta=1-sin theta
$$

$$
2sin^{2}theta+sin theta-1=1-sin theta+sin theta-1
$$

$$
2sin^{2}theta+sin theta-1=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin theta$ in our equation, so we can factor to find the values of $color{#4257b2}sin theta$.

$$
left(sin theta+1right)left(2sin theta-1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}sin theta$.

$$
sin theta+1=0 text{or} 2sin theta-1=0
$$

$$
sin theta=-1 text{or} sin theta=dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}sin theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}sin theta=-1$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(-1right)
$$

$$
theta=sin^{-1}left(-1right)
$$

$$
theta=dfrac{3pi}{2}=4.71 text{radians}
$$

Note that $color{#4257b2}dfrac{3pi}{2}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{3pi}{2}=-1$.

So the solution of the first case is $boxed{ theta=4.71 text{radians} }$

For $color{#4257b2}sin theta=dfrac{1}{2}$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(dfrac{1}{2}right)
$$

$$
theta=sin^{-1}left(dfrac{1}{2}right)
$$

$$
theta=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin theta=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin theta=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=dfrac{pi}{6} text{or} theta=pi-dfrac{pi}{6}
$$

$$
theta=dfrac{pi}{6}=0.52 text{radians} text{or} theta=dfrac{5pi}{6}=2.62 text{radians}
$$

So the solutions of the second case are $boxed{ theta=0.52 text{radians} } text{or} boxed{ theta=2.62 text{radians} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{0.52 text{radians}, 2.62 text{radians}, 4.71 text{radians}right}$

(c) We would like to solve the equation $color{#4257b2}cos^{2}theta=2+cos theta$ to the nearest hundredth where $color{#4257b2}0 leq theta leq 2pi$. First, we can subtract $color{#4257b2}cos theta+2$ from each side to make the right side equals zero.

$$
cos^{2}theta=2+cos theta
$$

$$
cos^{2}theta-cos theta-2=2+cos theta-cos theta-2
$$

$$
cos^{2}theta-cos theta-2=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}cos theta$ in our equation, so we can factor to find the values of $color{#4257b2}cos theta$.

$$
left(cos theta-1right)left(2cos theta+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos theta$.

$$
cos theta-1=0 text{or} 2cos theta+1=0
$$

$$
cos theta=1 text{or} cos theta=-dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}cos theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}cos theta=1$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(1right)
$$

$$
theta=cos^{-1}left(1right)
$$

$$
theta=0 text{or} theta=2pi=6.28 text{radians}
$$

Note that $color{#4257b2}0$ and $color{#4257b2}2pi$ are special angles which we know the values of the cosine function of them where $color{#4257b2}cos 0=cos 2pi=1$.

So the solutions of the first case are $boxed{ theta=0 } text{or} boxed{ theta=6.28 text{radians} }$

For $color{#4257b2}cos theta=-dfrac{1}{2}$

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(-dfrac{1}{2}right)
$$

$$
theta=cos^{-1}left(-dfrac{1}{2}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ to find the related acute angle.

$$
theta=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos theta=-dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos theta=-dfrac{1}{2}$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=pi-dfrac{pi}{3} text{or} theta=pi+dfrac{pi}{3}
$$

$$
theta=dfrac{2pi}{3}=2.09 text{radians} text{or} theta=dfrac{4pi}{3}=4.19 text{radians}
$$

So the solutions of the second case are $boxed{ theta=2.09 text{radians} } text{or} boxed{ theta=4.19 text{radians} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{0, 2.09 text{radians}, 4.19 text{radians}, 6.28 text{radians}right}$

(d) We would like to solve the equation $color{#4257b2}2sin^{2}theta+5sin theta-3=0$ to the nearest hundredth where $color{#4257b2}0 leq theta leq 2pi$. First, we note that our equation is a quadratic equation on the form $color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin theta$ in our equation, so we can factor to find the values of $color{#4257b2}sin theta$.

$$
2sin^{2}theta+5sin theta-3=0
$$

$$
left(sin theta+3right)left(2sin theta-1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}sin theta$.

$$
sin theta+3=0 text{or} 2sin theta-1=0
$$

$$
sin theta=-3 text{or} sin theta=dfrac{1}{2}
$$

But we know that $color{#4257b2}-1 leq sin theta leq 1$, so the solution $color{#4257b2}sin theta=-3$ is refused and the only solution is $color{#4257b2}sin theta=dfrac{1}{2}$.

$$
color{#4257b2}sin theta=dfrac{1}{2}
$$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(dfrac{1}{2}right)
$$

$$
theta=sin^{-1}left(dfrac{1}{2}right)
$$

$$
theta=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin theta=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin theta=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=dfrac{pi}{6} text{or} theta=pi-dfrac{pi}{6}
$$

$$
theta=dfrac{pi}{6}=0.52 text{radians} text{or} theta=dfrac{5pi}{6}=2.62 text{radians}
$$

So the solutions of the equation are $color{#4257b2}theta=left{0.52 text{radians}, 2.62 text{radians}right}$

(e) We would like to solve the equation $color{#4257b2}3tan^{2}theta-2tan theta=1$ to the nearest hundredth where $color{#4257b2}0 leq theta leq 2pi$. First, we can subtract $color{#4257b2}1$ from each side to make the right side equals zero.

$$
3tan^{2}theta-2tan theta=1
$$

$$
3tan^{2}theta-2tan theta-1=1-1
$$

$$
3tan^{2}theta-2tan theta-1=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}tan theta$ in our equation, so we can factor to find the values of $color{#4257b2}tan theta$.

$$
left(tan theta-1right)left(3tan theta+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}tan theta$.

$$
tan theta-1=0 text{or} 3tan theta+1=0
$$

$$
tan theta=1 text{or} tan theta=-dfrac{1}{3}
$$

Now we have two cases for $color{#4257b2}tan theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}tan theta=1$

$$
tan^{-1}left(tan thetaright)=tan^{-1}left(1right)
$$

$$
theta=tan^{-1}left(1right)
$$

$$
theta=dfrac{pi}{4}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan dfrac{pi}{4}=1$.

Now we found the related acute angle for the equation $color{#4257b2}tan theta=1$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan theta=1$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $3$ where the tangent ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=dfrac{pi}{4} text{or} theta=pi+dfrac{pi}{4}
$$

$$
theta=dfrac{pi}{4}=0.79 text{radians} text{or} theta=dfrac{5pi}{4}=3.93 text{radians}
$$

So the solutions of the first case are $boxed{ theta=0.79 text{radians} } text{or} boxed{ theta=3.93 text{radians} }$

For $color{#4257b2}tan theta=-dfrac{1}{3}$

$$
tan^{-1}left(tan thetaright)=tan^{-1}left(-dfrac{1}{3}right)
$$

$$
theta=tan^{-1}left(-dfrac{1}{3}right)
$$

Now we will calculate $color{#4257b2}tan^{-1}left(dfrac{1}{3}right)$ to find the related acute angle.

$$
theta=0.32 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}tan theta=-dfrac{1}{3}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan theta=-dfrac{1}{3}$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $4$ where the tangent ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=pi-0.32 text{or} theta=2pi-0.32
$$

$$
theta=2.82 text{radians} text{or} theta=5.96 text{radians}
$$

So the solutions of the second case are $boxed{ theta=2.82 text{radians} } text{or} boxed{ theta=5.96 text{radians} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{0.79 text{radians}, 2.82 text{radians}, 3.93 text{radians}, 5.96 text{radians}right}$

(f) We would like to solve the equation $color{#4257b2}12sin^{2}theta+sin theta-6=0$ to the nearest hundredth where $color{#4257b2}0 leq theta leq 2pi$. First, we note that our equation is a quadratic equation on the form $color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin theta$ in our equation, so we can factor to find the values of $color{#4257b2}sin theta$.

$$
12sin^{2}theta+sin theta-6=0
$$

$$
left(3sin theta-2right)left(4sin theta+3right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}sin theta$.

$$
3sin theta-2=0 text{or} 4sin theta+3=0
$$

$$
sin theta=dfrac{2}{3} text{or} sin theta=-dfrac{3}{4}
$$

Now we have two cases for $color{#4257b2}sin theta$, so we can solve each case to find the values of $color{#4257b2}theta$.

For $color{#4257b2}sin theta=dfrac{2}{3}$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(dfrac{2}{3}right)
$$

$$
theta=sin^{-1}left(dfrac{2}{3}right)
$$

$$
theta=0.73 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}sin theta=dfrac{2}{3}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin theta=dfrac{2}{3}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=0.73 text{or} theta=pi-0.73
$$

$$
theta=0.73 text{radians} text{or} theta=2.41 text{radians}
$$

So the solutions of the first case are $boxed{ theta=0.73 text{radians} } text{or} boxed{ theta=2.41 text{radians} }$

For $color{#4257b2}sin theta=-dfrac{3}{4}$

$$
sin^{-1}left(sin thetaright)=sin^{-1}left(-dfrac{3}{4}right)
$$

$$
theta=sin^{-1}left(-dfrac{3}{4}right)
$$

Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{3}{4}right)$ to find the related acute angle.

$$
theta=0.85 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}sin theta=-dfrac{3}{4}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin theta=-dfrac{3}{4}$ which means that it is negative, so the solutions in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=pi+0.85 text{or} theta=2pi-0.85
$$

$$
theta=3.99 text{radians} text{or} theta=5.44 text{radians}
$$

So the solutions of the second case are $boxed{ theta=3.99 text{radians} } text{or} boxed{ theta=5.44 text{radians} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}theta=left{0.73 text{radians}, 2.41 text{radians}, 3.99 text{radians}, 5.44 text{radians}right}$

$$
text{color{#c34632}(a) $theta=left{1.05 text{radians}, 3.14 text{radians}, 5.24 text{radians}right}$
\
\
(b) $theta=left{0.52 text{radians}, 2.62 text{radians}, 4.71 text{radians}right}$
\
\
(c) $theta=left{0, 2.09 text{radians}, 4.19 text{radians}, 6.28 text{radians}right}$
\
\
$(d) theta=left{0.52 text{radians}, 2.62 text{radians}right}$
\
\
$(e) theta=left{0.79 text{radians}, 2.82 text{radians}, 3.93 text{radians}, 5.96 text{radians}right}$
\
\
$(f) theta=left{0.73 text{radians}, 2.41 text{radians}, 3.99 text{radians}, 5.44 text{radians}right}$}
$$

(a) We would like to solve the equation $color{#4257b2}sec xcsc x-2csc x=0$ for
$color{#4257b2}0 leq x leq 2pi$. First, we note that all terms contain $color{#4257b2}csc x$, so we can take it as a common factor.

$$
sec xcsc x-2csc x=0
$$

$$
csc xleft(sec x-2right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
csc x=0 text{or} sec x-2=0
$$

$$
csc x=0 text{or} sec x=2
$$

But we know that $color{#4257b2}|csc x| geq 1$, so the solution $color{#4257b2}csc x=0$ is refused.

$$
sec x=2
$$

But we know that $color{#4257b2}sec x=dfrac{1}{cos x}$, so we can use this identity in our equation.

$$
dfrac{1}{cos x}=2
$$

$$
cos x=dfrac{1}{2}
$$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{3} text{or} x=2pi-dfrac{pi}{3}
$$

$$
x=dfrac{pi}{3} text{or} x=dfrac{5pi}{3}
$$

So the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{3}, dfrac{5pi}{3}right}$

(b) We would like to solve the equation $color{#4257b2}3sec^{2}x-4=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we can add $color{#4257b2}4$ to each side to make $color{#4257b2}sec^{2}x$ in the left side alone.

$$
3sec^{2}x-4=0
$$

$$
3sec^{2}x=4
$$

Now we can divide the two sides by $color{#4257b2}3$

$$
sec^{2}x=dfrac{4}{3}
$$

Now we can take the square root for each side to find the values of $color{#4257b2}sec x$

$$
sec x=pm sqrt{dfrac{4}{3}}=pm dfrac{2}{sqrt{3}}
$$

But we know that $color{#4257b2}sec x=dfrac{1}{cos x}$, so we can use this identity in our equation.

$$
dfrac{1}{cos x}=pm dfrac{2}{sqrt{3}}
$$

$$
cos x=pm dfrac{sqrt{3}}{2}
$$

Now we have two cases for $color{#4257b2}cos x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=dfrac{sqrt{3}}{2}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{sqrt{3}}{2}right)
$$

$$
x=cos^{-1}left(dfrac{sqrt{3}}{2}right)
$$

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{6}=dfrac{sqrt{3}}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{sqrt{3}}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{6} text{or} x=2pi-dfrac{pi}{6}=dfrac{11pi}{6}
$$

So the solutions of the first case are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{11pi}{6}right}$

For $color{#4257b2}cos x=-dfrac{sqrt{3}}{2}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{sqrt{3}}{2}right)
$$

$$
x=cos^{-1}left(-dfrac{sqrt{3}}{2}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{sqrt{3}}{2}right)$ to find the related acute angle.

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{6}=dfrac{sqrt{3}}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{sqrt{3}}{2}$ which means that it is positive, so the solutions in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{6} text{or} x=pi+dfrac{pi}{6}
$$

$$
x=dfrac{5pi}{6} text{or} x=dfrac{7pi}{6}
$$

So the solutions of the second case are $color{#4257b2}x=left{dfrac{5pi}{6}, dfrac{7pi}{6}right}$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{5pi}{6}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$

(c) We would like to solve the equation $color{#4257b2}2sin xsec x-2sqrt{3} sin x=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that all terms contain $color{#4257b2}2sin x$, so we can take it as a common factor.

$$
2sin xsec x-2sqrt{3} sin x=0
$$

$$
2sin xleft(sec x-sqrt{3}right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
2sin x=0 text{or} sec x-sqrt{3}=0
$$

$$
sin x=0 text{or} sec x=sqrt{3}
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=0$

$$
sin^{-1}left(sin xright)=sin^{-1}left(0right)
$$

$$
x=sin^{-1}left(0right)
$$

$$
x=0, pi text{or} x=2pi
$$

Note that $color{#4257b2}0, pi$ and $color{#4257b2}2pi$ are special angles which we know the values of the sine function for them where $color{#4257b2}sin 0=sin pi=sin 2pi=0$.

So the solutions of the first case are $boxed{ x=0, pi } text{or} boxed{ x=2pi }$

For $color{#4257b2}sec x=sqrt{3}$

Since we know that $color{#4257b2}sec x=dfrac{1}{cos x}$, so we can use this identity in our equation.

$$
dfrac{1}{cos x}=sqrt{3}
$$

$$
cos x=dfrac{1}{sqrt{3}}
$$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
x=cos^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
x=0.96 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{1}{sqrt{3}}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=0.96 text{or} x=2pi-0.96
$$

$$
x=0.96 text{radians} text{or} x=5.33 text{radians}
$$

So the solutions of the second case are $color{#4257b2}x=left{0.96 text{radians}, 5.33 text{radians}right}$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{0, pi, 2pi, 0.96 text{radians}, 5.33 text{radians}right}$

(d) We would like to solve the equation $color{#4257b2}2cot x+sec^{2}x=0$ for
$color{#4257b2}0 leq x leq 2pi$. First, we can use the Pythagorean identity $color{#4257b2}1+tan^{2}x=sec^{2}x$ to replace $color{#4257b2}sec^{2}x$ from our equation by $color{#4257b2}1+tan^{2}x$.

$$
2cot x+sec^{2}x=0
$$

$$
2cot x+1+tan^{2}x=0
$$

But we know that $color{#4257b2}cot x=dfrac{1}{tan x}$, so we can use this identity in our equation to simplify.

$$
dfrac{2}{tan x}+1+tan^{2}x=0
$$

Now we can multiply the two sides of the equation by $color{#4257b2}tan x$.

$$
tan xcdot dfrac{2}{tan x}+tan x+tan^{3}x=0
$$

$$
tan^{3}x+tan x+2=0
$$

$$
tan^{3}x-tan x+2tan x+2=0
$$

$$
left(tan^{3}x-tan xright)+left(2tan x+2right)=0
$$

$$
tan xleft(tan^{2}x-1right)+2left(tan x+1right)=0
$$

$$
tan xleft(tan x+1right)left(tan x-1right)+2left(tan x+1right)=0
$$

Note that we split $color{#4257b2}tan x$ to $color{#4257b2}-tan x+2tan x$ to factor the cubic equation. Now we note that our equation consists of two terms each of them contains the bracket $color{#4257b2}tan x+1$, so we can take it as a common factor.

$$
left(tan x+1right)left[tan xleft(tan x-1right)+2right]=0
$$

$$
left(tan x+1right)left(tan^{2}x-tan x+2right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
tan x+1=0 text{or} tan^{2}x-tan x+2=0
$$

$$
tan x=-1 text{or} tan^{2}x-tan x+2=0
$$

But we note that $color{#4257b2}tan^{2}x-tan x+2=0$ is a quadratic equation and if we calculated the term $color{#4257b2}sqrt{b^{2}-4ac}$ it will equal $color{#4257b2}sqrt{(-1)^{2}-4cdot 1cdot 4}=sqrt{-15}$, so the roots of this equation will be imaginary. As a result the solution of $color{#4257b2}tan^{2}x-tan x+2=0$ is refused.

$$
tan x=-1
$$

$$
tan^{-1}left(tan xright)=tan^{-1}left(-1right)
$$

$$
x=tan^{-1}left(-1right)
$$

Now we will calculate $color{#4257b2}tan^{-1}(1)$ to find the related acute angle.

$$
x=dfrac{pi}{4}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan dfrac{pi}{4}=1$.

Now we found the related acute angle for the equation $color{#4257b2}tan x=-1$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan x=-1$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $4$ where the tangent ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{4} text{or} x=2pi-dfrac{pi}{4}
$$

$$
x=dfrac{3pi}{4} text{or} x=dfrac{7pi}{4}
$$

So the solutions of the equation are $color{#4257b2}x=left{dfrac{3pi}{4}, dfrac{7pi}{4}right}$

(e) We would like to solve the equation $color{#4257b2}cot xcsc^{2}x=2cot x$ for
$color{#4257b2}0 leq x leq 2pi$. First, we can subtract $color{#4257b2}2cot x$ from each side to make the right side equals zero.

$$
cot xcsc^{2}x=2cot x
$$

$$
cot xcsc^{2}x-2cot x=0
$$

Now we note that all terms contain $color{#4257b2}cot x$, so we can take it as a common factor.

$$
cot xleft(csc^{2}x-2right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
cot x=0 text{or} csc^{2}x-2=0
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cot x=0$

$$
cot^{-1}left(cot xright)=cot^{-1}left(0right)
$$

$$
x=cot^{-1}left(0right)
$$

$$
x=dfrac{pi}{2} text{or} x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ and $color{#4257b2}dfrac{3pi}{2}$ are special angles which we know the values of the cotangent function for them where $color{#4257b2}cot dfrac{pi}{2}=cot dfrac{3pi}{2}=0$.

So the solutions of the first case are $boxed{ x=dfrac{pi}{2} } text{or} boxed{ x=dfrac{3pi}{2} }$.

For $color{#4257b2}csc^{2}x-2=0$

$$
csc^{2}x=2
$$

Now we can take the square root for each side to find the values of $color{#4257b2}csc x$.

$$
csc x=pm sqrt{2}
$$

But we know that $color{#4257b2}csc x=dfrac{1}{sin x}$, so we can use this identity in our equation.

$$
dfrac{1}{sin x}=pm sqrt{2}
$$

$$
sin x=pm dfrac{1}{sqrt{2}}
$$

Now we have two cases for $color{#4257b2}sin x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=dfrac{1}{sqrt{2}}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
x=sin^{-1}left(dfrac{1}{sqrt{2}}right)
$$

$$
x=dfrac{pi}{4}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{4}=dfrac{1}{sqrt{2}}$

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=dfrac{1}{sqrt{2}}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{4} text{or} x=pi-dfrac{pi}{4}=dfrac{3pi}{4}
$$

So the solutions of the second case are $color{#4257b2}x=left{dfrac{pi}{4}, dfrac{3pi}{4}right}$

For $color{#4257b2}sin x=-dfrac{1}{sqrt{2}}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

$$
x=sin^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{sqrt{2}}right)$ to find the related acute angle.

$$
x=dfrac{pi}{4}
$$

Now we found the related acute angle for the equation $color{#4257b2}sin x=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi+dfrac{pi}{4}=dfrac{5pi}{4} text{or} x=2pi-dfrac{pi}{4}=dfrac{7pi}{4}
$$

So the solutions of the third case are $color{#4257b2}x=left{dfrac{5pi}{4}, dfrac{7pi}{4}right}$

Now we found the solutions of the three cases, so the solutions of the equation are $color{#4257b2}xleft{dfrac{pi}{4}, dfrac{pi}{2}, dfrac{3pi}{4}, dfrac{5pi}{4}, dfrac{3pi}{2}, dfrac{7pi}{4}right}$

(f) We would like to solve the equation $color{#4257b2}3tan^{3}x-tan x=0$ for
$color{#4257b2}0 leq x leq 2pi$. First, we note that all terms contain $color{#4257b2}tan x$, so we can take it as a common factor.

$$
3tan^{3}x-tan x=0
$$

$$
tan xleft(3tan^{2}x-1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}x$.

$$
tan x=0 text{or} 3tan^{2}x-1=0
$$

$$
tan x=0 text{or} tan^{2}x=dfrac{1}{3}
$$

Now we have two cases, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}tan x=0$

$$
tan^{-1}left(tan xright)=tan^{-1}left(0right)
$$

$$
x=tan^{-1}left(0right)
$$

$$
x=0, pi text{or} x=2pi
$$

Note that $color{#4257b2}0, pi$ and $color{#4257b2}2pi$ are special angles which we know the values of the tangent function for them where $color{#4257b2}tan 0=tan pi=tan 2pi=0$.

So the solutions of the first case are $boxed{ x=0, pi } text{or} boxed{ x=2pi }$.

For $color{#4257b2}tan^{2}x=dfrac{1}{3}$

We can take the square root for each side to find the values of $color{#4257b2}tan x$.

$$
tan x=pm sqrt{dfrac{1}{3}}=pm dfrac{1}{sqrt{3}}
$$

Now we have two cases for $color{#4257b2}tan x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$

$$
tan^{-1}left(tan xright)=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
x=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan dfrac{pi}{6}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $3$ where the tangent ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{6} text{or} x=pi+dfrac{pi}{6}
$$

$$
x=dfrac{pi}{6} text{or} x=dfrac{7pi}{6}
$$

So the solutions of the second case are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{7pi}{6}right}$

For $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$

$$
tan^{-1}left(tan xright)=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

$$
x=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

Now we will calculate $color{#4257b2}tan^{-1}left(dfrac{1}{sqrt{3}}right)$ to find the related acute angle.

$$
x=dfrac{pi}{6}
$$

Now we found the related acute angle for the equation $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $4$ where the tangent ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{6}=dfrac{5pi}{6} text{or} x=2pi-dfrac{pi}{6}=dfrac{11pi}{6}
$$

So the solutions of the third case are $color{#4257b2}x=left{dfrac{5pi}{6}, dfrac{11pi}{6}right}$

Now we found the solutions of the three cases, so the solutions of the equation are $color{#4257b2}xleft{0, dfrac{pi}{6}, dfrac{5pi}{6}, pi, dfrac{7pi}{6}, dfrac{11pi}{6}, 2piright}$

$$
text{color{#c34632}$(a) x=left{dfrac{pi}{3}, dfrac{5pi}{3}right}$ $(d) x=left{dfrac{3pi}{4}, dfrac{7pi}{4}right}$
\
\
\
color{#c34632}$(b) x=left{dfrac{pi}{6}, dfrac{5pi}{6}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$ $(e) xleft{dfrac{pi}{4}, dfrac{pi}{2}, dfrac{3pi}{4}, dfrac{5pi}{4}, dfrac{3pi}{2}, dfrac{7pi}{4}right}$
\
\
\
color{#c34632}$(c) x=left{0, pi, 2pi, 0.96 text{radians}, 5.33 text{radians}right}$ $(f) xleft{0, dfrac{pi}{6}, dfrac{5pi}{6}, pi, dfrac{7pi}{6}, dfrac{11pi}{6}, 2piright}$}
$$

(a) We would like to solve the equation $color{#4257b2}5cos 2x-cos x+3=0$ where $color{#4257b2}0 leq x leq 2pi$. First, we can use the double angle formula of the cosine function where $color{#4257b2}cos 2x=2cos^{2}x-1$

$$
5cos 2x-cos x+3=0
$$

$$
5left(2cos^{2}x-1right)-cos x+3=0
$$

$$
10cos^{2}x-5-cos x+3=0
$$

$$
10cos^{2}x-cos x-2=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}cos x$ in our equation, so we can factor to find the values of $color{#4257b2}cos x$.

$$
left(5cos x+2right)left(2cos x-1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos x$.

$$
5cos x+2=0 text{or} 2cos x-1=0
$$

$$
cos x=-dfrac{2}{5} text{or} cos x=dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}cos x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=-dfrac{2}{5}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{2}{5}right)
$$

$$
x=cos^{-1}left(-dfrac{2}{5}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{2}{5}right)$ to determine the related acute angle.

$$
x=1.16 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{2}{5}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{2}{5}$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-1.16 text{or} x=pi+1.16
$$

$$
x=1.98 text{radians} text{or} x=4.3 text{radians}
$$

So the solutions of the first case are $boxed{ x=1.98 text{radians} } text{or} boxed{ x=4.3 text{radians} }$

For $color{#4257b2}cos x=dfrac{1}{2}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=cos^{-1}left(dfrac{1}{2}right)
$$

$$
x=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{3} text{or} x=2pi-dfrac{pi}{3}
$$

$$
x=dfrac{pi}{3} text{or} x=dfrac{5pi}{3}
$$

So the solutions of the second case are $boxed{ x=dfrac{pi}{3} } text{or} boxed{ x=dfrac{5pi}{3} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{3}, dfrac{5pi}{3}, 1.98 text{radians}, 4.3 text{radians}right}$

(b) We would like to solve the equation $color{#4257b2}10cos 2x-8cos x+1=0$ where $color{#4257b2}0 leq x leq 2pi$. First, we can use the double angle formula of the cosine function where $color{#4257b2}cos 2x=2cos^{2}x-1$

$$
10cos 2x-8cos x+1=0
$$

$$
10left(2cos^{2}x-1right)-8cos x+1=0
$$

$$
20cos^{2}x-10-8cos x+1=0
$$

$$
20cos^{2}x-8cos x-9=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}cos x$ in our equation, so we can factor to find the values of $color{#4257b2}cos x$.

$$
left(10cos x-9right)left(2cos x+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos x$.

$$
10cos x-9=0 text{or} 2cos x+1=0
$$

$$
cos x=dfrac{9}{10} text{or} cos x=-dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}cos x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=dfrac{9}{10}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{9}{10}right)
$$

$$
x=cos^{-1}left(dfrac{9}{10}right)
$$

$$
x=0.45 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{9}{10}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{9}{10}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=0.45 text{or} x=2pi-0.45
$$

$$
x=0.45 text{radians} text{or} x=5.83 text{radians}
$$

So the solutions of the first case are $boxed{ x=0.45 text{radians} } text{or} boxed{ x=5.83 text{radians} }$

For $color{#4257b2}cos x=-dfrac{1}{2}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{1}{2}right)
$$

$$
x=cos^{-1}left(-dfrac{1}{2}right)
$$

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ to determine the related acute angle.

$$
x=dfrac{pi}{3}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{1}{2}$ which means that it is negative, so the solutions in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{3} text{or} x=pi+dfrac{pi}{3}
$$

$$
x=dfrac{2pi}{3} text{or} x=dfrac{4pi}{3}
$$

So the solutions of the second case are $boxed{ x=dfrac{2pi}{3} } text{or} boxed{ x=dfrac{4pi}{3} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{2pi}{3}, dfrac{4pi}{3}, 0.45 text{radians}, 5.83 text{radians}right}$

(c) We would like to solve the equation $color{#4257b2}4cos 2x+10sin x-7=0$ where $color{#4257b2}0 leq x leq 2pi$. First, we can use the double angle formula of the cosine function where $color{#4257b2}cos 2x=1-2sin^{2}x$

$$
4cos 2x+10sin x-7=0
$$

$$
4left(1-2sin^{2}xright)+10sin x-7=0
$$

$$
4-8sin^{2}x+10sin x-7=0
$$

$$
-8sin^{2}x+10sin x-3=0
$$

$$
8sin^{2}x-10sin x+3=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin x$ in our equation, so we can factor to find the values of $color{#4257b2}sin x$.

$$
left(4sin x-3right)left(2sin x-1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}sin x$.

$$
4sin x-3=0 text{or} 2sin x-1=0
$$

$$
sin x=dfrac{3}{4} text{or} sin x=dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}sin x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=dfrac{3}{4}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{3}{4}right)
$$

$$
x=sin^{-1}left(dfrac{3}{4}right)
$$

$$
x=0.85 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{3}{4}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=dfrac{3}{4}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=0.85 text{or} x=pi-0.85
$$

$$
x=0.85 text{radians} text{or} x=2.29 text{radians}
$$

So the solutions of the first case are $boxed{ x=0.85 text{radians} } text{or} boxed{ x=2.29 text{radians} }$

For $color{#4257b2}sin x=dfrac{1}{2}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(dfrac{1}{2}right)
$$

$$
x=sin^{-1}left(dfrac{1}{2}right)
$$

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=dfrac{1}{2}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=dfrac{pi}{6} text{or} x=pi-dfrac{pi}{6}
$$

$$
x=dfrac{pi}{6} text{or} x=dfrac{5pi}{6}
$$

So the solutions of the second case are $boxed{ x=dfrac{pi}{6} } text{or} boxed{ x=dfrac{5pi}{6} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{6}, dfrac{5pi}{6}, 0.85 text{radians}, 2.29 text{radians}right}$

(d) We would like to solve the equation $color{#4257b2}-2cos 2x=2sin x$ where $color{#4257b2}0 leq x leq 2pi$. First, we can subtract $color{#4257b2}2sin x$ from each side to make the right side equals zero.

$$
-2cos 2x=2sin x
$$

$$
-2cos 2x-2sin x=0
$$

Now we can divide the two sides by $2$.

$$
-cos 2x-sin x=0
$$

Now we can use the double angle formula of the cosine function where $color{#4257b2}cos 2x=1-2sin^{2}x$.

$$
-left(1-2sin^{2}xright)-sin x=0
$$

$$
-1+2sin^{2}x-sin x=0
$$

$$
2sin^{2}x-sin x-1=0
$$

Now we note that our equation is a quadratic equation on the form
$color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ here is considered to be $color{#4257b2}sin x$ in our equation, so we can factor to find the values of $color{#4257b2}sin x$.

$$
left(sin x-1right)left(2sin x+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}sin x$.

$$
sin x-1=0 text{or} 2sin x+1=0
$$

$$
sin x=1 text{or} sin x=-dfrac{1}{2}
$$

Now we have two cases for $color{#4257b2}sin x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}sin x=1$

$$
sin^{-1}left(sin xright)=sin^{-1}left(1right)
$$

$$
x=sin^{-1}left(1right)
$$

$$
x=dfrac{pi}{2}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{2}=1$.

So the solution of the first case is $boxed{ x=dfrac{pi}{2} }$

For $color{#4257b2}sin x=-dfrac{1}{2}$

$$
sin^{-1}left(sin xright)=sin^{-1}left(-dfrac{1}{2}right)
$$

$$
x=sin^{-1}left(-dfrac{1}{2}right)
$$

Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{2}right)$ to find the related acute angle.

$$
x=dfrac{pi}{6}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

Now we found the related acute angle for the equation $color{#4257b2}sin x=-dfrac{1}{2}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=-dfrac{1}{2}$ which means that it is negative, so the solutions in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi+dfrac{pi}{6} text{or} x=2pi-dfrac{pi}{6}
$$

$$
x=dfrac{7pi}{6} text{or} x=dfrac{11pi}{6}
$$

So the solutions of the second case are $boxed{ x=dfrac{7pi}{6} } text{or} boxed{ x=dfrac{11pi}{6} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{2}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$

$$
text{color{#c34632}(a) $x=left{dfrac{pi}{3}, dfrac{5pi}{3}, 1.98 text{radians}, 4.3 text{radians}right}$
\
\
\
(b) $x=left{dfrac{2pi}{3}, dfrac{4pi}{3}, 0.45 text{radians}, 5.83 text{radians}right}$
\
\
\
(c) $x=left{dfrac{pi}{6}, dfrac{5pi}{6}, 0.85 text{radians}, 2.29 text{radians}right}$
\
\
\
$(d) x=left{dfrac{pi}{2}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$}
$$

We would like to solve the equation $color{#4257b2}8sin^{2}x-8sin x+1=0$ for $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we have a quadratic equation on the form $color{#4257b2}a y^{2}+b y+c=0$ where $color{#4257b2}y$ is considered to be $color{#4257b2}sin x$ in our equation, so we can use the quadratic formula to find the values of $color{#4257b2}sin x$.

$$
8sin^{2}x-8sin x+1=0
$$

$$
sin x=dfrac{-b pm sqrt{b^{2}-4 a c}}{2a}
$$

$$
sin x=dfrac{-(-8) pm sqrt{(-8)^{2}-4cdot 8cdot 1}}{2cdot 8}
$$

Note that our equation $color{#4257b2}8sin^{2}x-8sin x+1=0$ is on the form $color{#4257b2}a y^{2}+b y+c=0$, so we substituted the values $color{#4257b2}a=8, b=-8$ and $color{#4257b2}c=1$ in the quadratic formula.

$$
sin x=dfrac{8 pm sqrt{64-32}}{16}
$$

$$
sin x=dfrac{8 pm sqrt{32}}{16}
$$

$$
sin x=dfrac{8 pm 4sqrt{2}}{16}
$$

$$
sin x=dfrac{2 pm sqrt{2}}{4}
$$

$$
sin x=dfrac{2+sqrt{2}}{4} text{or} sin x=dfrac{2-sqrt{2}}{4}
$$

$$
sin x=0.85 text{or} sin x=0.15
$$

Now we have two cases for $color{#4257b2}sin x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $sin x=0.85$

We can take $color{#4257b2}sin^{-1}$ for each side to find the related acute angle.

$$
sin^{-1}left(sin xright)=sin^{-1}left(0.85right)
$$

$$
x=sin^{-1}left(0.85right)
$$

$$
x=1.02 text{radian}
$$

Now we found the related acute angle for the equation $color{#4257b2}sin x=0.85$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=0.85$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=1.02 text{or} x=pi-1.02
$$

$$
x=1.02 text{radians} text{or} x=2.12 text{radians}
$$

So the solutions of the first case are $boxed{ x=1.02 text{radians} } text{or} boxed{ x=2.12 text{radians} }$

For $sin x=0.15$

We can take $color{#4257b2}sin^{-1}$ for each side to find the related acute angle.

$$
sin^{-1}left(sin xright)=sin^{-1}left(0.15right)
$$

$$
x=sin^{-1}left(0.15right)
$$

$$
x=0.15 text{radian}
$$

Now we found the related acute angle for the equation $color{#4257b2}sin x=0.15$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}sin x=0.15$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $2$ where the sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=0.15 text{or} x=pi-0.15
$$

$$
x=0.15 text{radians} text{or} x=2.99 text{radians}
$$

So the solutions of the second case are $boxed{ x=0.15 text{radians} } text{or} boxed{ x=2.99 text{radians} }$

Now we found the solutions of each case, so the solutions of the equation are $color{#4257b2}x=left{0.15 text{radians}, 1.02 text{radians}, 2.12 text{radians}, 2.99 text{radians}right}$

$$
color{#c34632}x=left{0.15 text{radians}, 1.02 text{radians}, 2.12 text{radians}, 2.99 text{radians}right}
$$

We would like to find the values $color{#4257b2}b$ and $color{#4257b2}c$ in the trigonometric equation $color{#4257b2}cot^{2}x-b cot x+c=0$ if we know that this equation has the solutions $color{#4257b2}dfrac{pi}{6}, dfrac{pi}{4}, dfrac{7pi}{6}$ and $color{#4257b2}dfrac{5pi}{4}$. First, Since we know the solutions of the equation, so we can substitute the first two solutions in the equation and find two equations in the two variables $color{#4257b2}b$ and $color{#4257b2}c$ and then solve them to find the values of them.

For the solution $color{#4257b2}dfrac{pi}{6}$

$$
cot^{2}x-b cot x+c=0
$$

$$
cot^{2}dfrac{pi}{6}-b cot dfrac{pi}{6}+c=0
$$

$$
left(sqrt{3}right)^{2}-b cdot sqrt{3}+c=0
$$

Note that $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the cotangent function of it where $color{#4257b2}cot dfrac{pi}{6}=sqrt{3}$.

$$
3-sqrt{3} b+c=0
$$

$$
c-sqrt{3} b=-3 (1)
$$

Now we found the first equation which relates the two variables $color{#4257b2}b$ and $color{#4257b2}c$, so the next step is to substitute the solution $color{#4257b2}dfrac{pi}{4}$ in the trigonometric equation to find the second equation.

For the solution $color{#4257b2}dfrac{pi}{4}$

$$
cot^{2}x-b cot x+c=0
$$

$$
cot^{2}dfrac{pi}{4}-b cot dfrac{pi}{4}+c=0
$$

$$
1^{2}-b cdot 1+c=0
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the cotangent function of it where $color{#4257b2}cot dfrac{pi}{4}=1$.

$$
1- b+c=0
$$

$$
b=c+1 (2)
$$

Note that we substituted the first two solutions because the other two solutions will give us the same equations because the cotangent function has the same value for $color{#4257b2}dfrac{pi}{6}$ and $color{#4257b2}dfrac{7pi}{6}$ and also has the same value for $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{5pi}{4}$.

Now we found two equations in the two variables $color{#4257b2}b$ and $color{#4257b2}c$, so we can solve these equations to find the values of $color{#4257b2}b$ and $color{#4257b2}c$. To solve these equations we can substitute the equation (2) in the equation (1) to find the value of $color{#4257b2}c$.

$$
c-sqrt{3} b=-3 (1), b=c+1 (2)
$$

$$
c-sqrt{3} left(c+1right)=-3
$$

$$
c-sqrt{3} c-sqrt{3}=-3
$$

$$
c-sqrt{3} c=-3+sqrt{3}
$$

$$
cleft(1-sqrt{3}right)=-3+sqrt{3}
$$

$$
c=dfrac{-3+sqrt{3}}{1-sqrt{3}}
$$

$$
boxed{ c=sqrt{3} }
$$

Now we can substitute the value of $color{#4257b2}c$ in equation (2) to find the value of $color{#4257b2}b$.

$$
b=c+1
$$

$$
boxed{ b=sqrt{3}+1 }
$$

$$text{color{#c34632}b=sqrt{3}+1 {color{Black$text{and}} $c=sqrt{3}$}}$

We would like to find the value of $color{#4257b2}c$ if we have the graph of the quadratic trigonometric equation $color{#4257b2}sin^{2}x-c=0$ as shown.

First, we note that the graph is for a quadratic sine function but shifted down by the value of $color{#4257b2}c$, so we can graph the function $color{#4257b2}sin^{2}x$ and then compare the two graphs with each other to know the value which is shifted from the graph of $color{#4257b2}sin^{2}x$ to the graph $color{#4257b2}sin^{2}x-c$.

Now by comparing the two graph with each other we find that the graph of $color{#4257b2}sin^{2}x-c$ is shifted down by the value $color{#4257b2}0.5$, so this will be the value of $color{#4257b2}c$ because we know that the graph of the function $color{#4257b2}sin^{2}x-c$ is the same of the graph of the function $color{#4257b2}sin^{2}x$ but the only difference that it is shifted down by the value $color{#4257b2}c$.

$$
boxed{ c=0.5 }
$$

Another solution for this problem:

Since we know the graph of the function $color{#4257b2}sin^{2}x-c$, so we can find the value of the function $color{#4257b2}sin^{2}x-c$ at $color{#4257b2}x=0$ from the graph and then substitute in the function $color{#4257b2}y=sin^{2}x-c$ to find the value of $color{#4257b2}c$.

We note from the graph that at $color{#4257b2}x=0$, then $color{#4257b2}y=-0.5$, so we can substitute these values in the function $color{#4257b2}y=sin^{2}x-c$ to find the value of $color{#4257b2}c$.

$$
y=sin^{2}x-c
$$

$$
-0.5=sin^{2}0-c
$$

$$
-0.5=0^{2}-c
$$

Note that $color{#4257b2}0$ is a special angle which we know the value of the sine function of it where $color{#4257b2}sin 0=0$.

$$
-0.5=-c
$$

$$
boxed{ c=0.5 }
$$

So we found that the value of $color{#4257b2}c$ equals $color{#4257b2}0.5$ and this is the same result of the first solution.

$$
color{#c34632}c=0.5
$$

We would like to solve the equation $color{#4257b2}6sin^{2}x=17cos x+11$ for $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we know that the Pythagorean identity is $color{#4257b2}sin^{2}x+cos^{2}x=1$, so $color{#4257b2}sin^{2}x=1-cos^{2}x$ and we can use it to replace $color{#4257b2}sin^{2}x$ in the left side by $color{#4257b2}1-cos^{2}x$.

$$
6sin^{2}x=17cos x+11
$$

$$
6left(1-cos^{2}xright)=17cos x+11
$$

$$
6-6cos^{2}x=17cos x+11
$$

Now we can add $color{#4257b2}6cos^{2}x-6$ to each side to make the left side equals zero.

$$
6-6cos^{2}x+6cos^{2}x-6=17cos x+11+6cos^{2}x-6
$$

$$
0=6cos^{2}x+17cos x+5
$$

Now we have a quadratic equation on the form $color{#4257b2}a y^{2}+b y+c=0$ where $color{#4257b2}y$ is considered to be $color{#4257b2}cos x$ in our equation, so we can factor to find the values of $color{#4257b2}cos x$.

$$
6cos^{2}x+17cos x+5=0
$$

$$
left(2cos x+5right)left(3cos x+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos x$.

$$
2cos x+5=0 text{or} 3cos x+1=0
$$

$$
cos x=-dfrac{5}{2} text{or} cos x=-dfrac{1}{3}
$$

But we know that $color{#4257b2}-1 leq cos x leq 1$, so the solution $color{#4257b2}cos x=-dfrac{5}{2}$ is refused.

$$
cos x=-dfrac{1}{3}
$$

Now we can take $color{#4257b2}cos^{-1}$ for each side to find the related acute angle.

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{1}{3}right)
$$

$$
x=cos^{-1}left(-dfrac{1}{3}right)
$$

Now we can use the calculator to determine $color{#4257b2}cos^{-1}left(dfrac{1}{3}right)$ to find the related acute angle.

$$
x=1.23 text{radian}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{1}{3}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{1}{3}$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-1.23 text{or} x=pi+1.23
$$

$$
boxed{ x=1.91 text{radians} } text{or} boxed{ x=4.37 text{radians} }
$$

$$
color{#c34632}x=1.91 text{radians} {color{Black}text{or}} x=4.37 text{radians}
$$

(a) We would like to solve the equation
$color{#4257b2}sin^{2}x-sqrt{2} cos x=cos^{2}x+sqrt{2} cos x+2$ for $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we know that the Pythagorean identity is $color{#4257b2}sin^{2}x+cos^{2}x=1$, so $color{#4257b2}sin^{2}x=1-cos^{2}x$ and we can use it to replace $color{#4257b2}sin^{2}x$ in the left side by $color{#4257b2}1-cos^{2}x$.

$$
sin^{2}x-sqrt{2} cos x=cos^{2}x+sqrt{2} cos x+2
$$

$$
1-cos^{2}x-sqrt{2} cos x=cos^{2}x+sqrt{2} cos x+2
$$

Now we can add $color{#4257b2}cos^{2}x+sqrt{2} cos x-1$ to each side to make the left side equals zero.

$$
1-cos^{2}x-sqrt{2} cos x+cos^{2}x+sqrt{2} cos x-1=cos^{2}x+sqrt{2} cos x+2+cos^{2}x+sqrt{2} cos x-1
$$

$$
0=2cos^{2}x+2sqrt{2} cos x+1
$$

Now we have a quadratic equation on the form $color{#4257b2}a y^{2}+b y+c=0$ where $color{#4257b2}y$ is considered to be $color{#4257b2}cos x$ in our equation, so we can factor to find the values of $color{#4257b2}cos x$.

$$
2cos^{2}x+2sqrt{2} cos x+1=0
$$

$$
left(sqrt{2} cos x+1right)left(sqrt{2} cos x+1right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos x$.

$$
sqrt{2} cos x+1=0 text{or} sqrt{2} cos x+1=0
$$

$$
cos x=-dfrac{1}{sqrt{2}} text{or} cos x=-dfrac{1}{sqrt{2}}
$$

Now we note that we have two similar equations, so we can solve one of them to find the values of $color{#4257b2}x$

$$
cos x=-dfrac{1}{sqrt{2}}
$$

Now we can take $color{#4257b2}cos^{-1}$ for each side to find the related acute angle.

$$
cos^{-1}left(cos xright)=cos^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

$$
x=cos^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

Now we can use the calculator to determine $color{#4257b2}cos^{-1}left(dfrac{1}{sqrt{2}}right)$ to find the related acute angle.

$$
x=dfrac{pi}{4}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{pi}{4}=dfrac{1}{sqrt{2}}$.

Now we found the related acute angle for the equation $color{#4257b2}cos x=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $3$ where the cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=pi-dfrac{pi}{4} text{or} x=pi+dfrac{pi}{4}
$$

$$
boxed{ x=dfrac{3pi}{4} } text{or} boxed{ x=dfrac{5pi}{4} }
$$

(b) We would like to write a general solution for the equation in part a). First, we know that the period of the cosine function is $color{#4257b2}2pi$, so we can add $color{#4257b2}2pi$ to the two sides of the solutions in part a) to find the general solution of this equation.

$$
x=dfrac{3pi}{4} text{or} x=dfrac{5pi}{4}
$$

$$
x=dfrac{3pi}{4}+2pi text{or} x=dfrac{5pi}{4}+2pi
$$

So the general solution of the equation is $boxed{ x=dfrac{3pi}{4}+2pi } text{or} boxed{ x=dfrac{5pi}{4}+2pi }$

$$
text{color{#c34632}(a) $x=dfrac{3pi}{4} {color{Black}text{or}} x=dfrac{5pi}{4}$
\
\
\
Large{color{#c34632}(b) $x=dfrac{3pi}{4}+2pi {color{Black}text{or}} x=dfrac{5pi}{4}+2pi$}}
$$

Since we know that any quadratic trigonometric equation can be written on the form $color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ is considered to be any trigonometric function, so to solve the quadratic trigonometric equation on this form we can use the quadratic formula $color{#4257b2}x=dfrac{-b pm sqrt{b^{2}-4ac}}{2a}$. Form this formula we note that the term under the square root is $color{#4257b2}b^{2}-4ac$ and this term has three cases:

The first case for $color{#4257b2}b^{2}-4ac > 0$

In this case there will be two real solutions for $color{#4257b2}x$ which is considered to be the trigonometric function. As a result in this case there will be two solutions for the quadratic trigonometric equation. For example:
$$
color{#4257b2}2sin^{2}theta-sin theta-1=0
$$

The solutions will be $color{#4257b2}sin theta=1 text{or} sin theta=-dfrac{1}{2}$, so there are two possible solutions for $color{#4257b2}sin theta$.

Also, in the first case it is possible that one of the two solutions is greater than $color{#4257b2}1$ or less than $color{#4257b2}-1$ and if the trigonometric function in this case is sine or cosine, so this solution will be refused because we know that $color{#4257b2}-1 leq sin theta (cos theta) leq 1$. As a result in this case there will be one solution for the quadratic trigonometric equation. For example:
$$
color{#4257b2}sin^{2}theta-sin theta-2=0
$$

The solutions will be $color{#4257b2}sin theta=-1 text{or} sin theta=2$, so there is one possible solution for $color{#4257b2}sin theta$ because the solution $color{#4257b2}sin theta=2$ is refused.

The second case for $color{#4257b2}b^{2}-4ac = 0$

In this case there will be two similar real solutions for $color{#4257b2}x$ which is considered to be the trigonometric function. As a result in this case there will be one solution for the quadratic trigonometric equation. For example:
$$
color{#4257b2}4sin^{2}theta-4sin theta+1=0
$$

The solutions will be $color{#4257b2}sin theta=dfrac{1}{2} text{or} sin theta=dfrac{1}{2}$, so there are one possible solution for $color{#4257b2}sin theta$.

Also, in the second case it is possible that these two similar solutions are greater than $color{#4257b2}1$ or less than $color{#4257b2}-1$ and if the trigonometric function in this case is sine or cosine, so this solution will be refused because we know that $color{#4257b2}-1 leq sin theta (cos theta) leq 1$. As a result in this case there will be no solution for the quadratic trigonometric equation. For example:
$$
color{#4257b2}sin^{2}theta-4sin theta+4=0
$$

The solutions will be $color{#4257b2}sin theta=2 text{or} sin theta=2$, so there is no possible solution for $color{#4257b2}sin theta$ because the solution $color{#4257b2}sin theta=2$ is refused.

The third case for $color{#4257b2}b^{2}-4ac < 0$

In this case there will be two imaginary solutions for $color{#4257b2}x$ which is considered to be the trigonometric function. As a result in this case there will be no solution for the quadratic trigonometric equation. For example:
$$
color{#4257b2}sin^{2}theta-2sin theta+4=0
$$

The solutions will be $color{#4257b2}sin theta=1+sqrt{3} i text{or} sin theta=1-sqrt{3} i$, so there is no possible solution for $color{#4257b2}sin theta$ because the trigonometric function doesn't equal imaginary solution.

So we proved with examples that it is possible to have different numbers of solutions for quadratic trigonometric equations.

$$
text{color{#c34632}It is possible to have different numbers of solutions for quadratic trigonometric equations}
$$

We would like to determine the values of $color{#4257b2}a$ in the interval $color{#4257b2}0 leq a leq 2pi$ if we know that the function $color{#4257b2}f(x)=dfrac{tan x}{1-tan x}-dfrac{cot x}{1-cot x}$ can be written as
$color{#4257b2}f(x)=tanleft(x+aright)$. First, we will simplify the function
$color{#4257b2}f(x)=dfrac{tan x}{1-tan x}-dfrac{cot x}{1-cot x}$ to be on the form of the addition formula for the tangent function.

$$
f(x)=dfrac{tan x}{1-tan x}-dfrac{cot x}{1-cot x}
$$

$$
f(x)=dfrac{tan x}{1-tan x}-dfrac{dfrac{1}{tan x}}{1-dfrac{1}{tan x}}
$$

$$
f(x)=dfrac{tan x}{1-tan x}-dfrac{1}{tan xleft(1-dfrac{1}{tan x}right)}
$$

$$
f(x)=dfrac{tan x}{1-tan x}-dfrac{1}{tan x-1}
$$

$$
f(x)=dfrac{tan x}{1-tan x}+dfrac{1}{1-tan x}
$$

Note that we used the identity $color{#4257b2}cot x=dfrac{1}{tan x}$ to simplify our function. Now we can unify the denominators to the common denominator $color{#4257b2}1-tan x$.

$$
f(x)=dfrac{tan x+1}{1-tan x}
$$

$$
f(x)=dfrac{1+tan x}{1-tan x}
$$

$$
f(x)=dfrac{1+tan x}{1-(1)cdot tan x}
$$

But we know that $color{#4257b2}tan dfrac{pi}{4}=tan dfrac{5pi}{4}=1$, so we can replace $color{#4257b2}1$ from the numerator and denominator by $color{#4257b2}tan dfrac{pi}{4}$ and also by $color{#4257b2}tan dfrac{5pi}{4}$

$$
f(x)=dfrac{1+tan x}{1-(1)cdot tan x}
$$

$$
f(x)=dfrac{tan dfrac{pi}{4}+tan x}{1-tan dfrac{pi}{4} tan x} text{or} f(x)=dfrac{tan dfrac{5pi}{4}+tan x}{1-tan dfrac{5pi}{4} tan x}
$$

Note that we made the final step to make our function on the form of the addition formula for the tangent function. Now we note that our function is on the form of the addition formula for the tangent function
$color{#4257b2}tan (a+b)=dfrac{tan a+tan b}{1-tan tan b}$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}x$ in the first function and considered to be $color{#4257b2}dfrac{5pi}{4}$ and $color{#4257b2}x$ in the second function. As a result we can use that formula to simplify our function.

$$
f(x)=dfrac{tan dfrac{pi}{4}+tan x}{1-tan dfrac{pi}{4} tan x} text{or} f(x)=dfrac{tan dfrac{5pi}{4}+tan x}{1-tan dfrac{5pi}{4} tan x}
$$

$$
f(x)=tan left(dfrac{pi}{4}+xright) text{or} f(x)=tan left(dfrac{5pi}{4}+xright)
$$

$$
f(x)=tan left(x+dfrac{pi}{4}right) text{or} f(x)=tan left(x+dfrac{5pi}{4}right)
$$

Now we note that our function is on the form $color{#4257b2}f(x)=tan (x+a)$ where $color{#4257b2}a$ is considered to be $color{#4257b2}dfrac{pi}{4}$ in the first function and considered to be $color{#4257b2}dfrac{5pi}{4}$ in the second function.

So the values of $color{#4257b2}a$ are $boxed{ a=dfrac{pi}{4} } text{or} boxed{ a=dfrac{5pi}{4} }$

$$text{color{#c34632}a=dfrac{pi}{4} {color{Black$text{or}} $a=dfrac{5pi}{4}$}}$

We would like to solve the equation $color{#4257b2}2cos 3x+cos 2x+1=0$. First, we can rewrite $color{#4257b2}cos 3x=cos (2x+x)$ and then use the addition formula for the cosine function where $color{#4257b2}cos (a+b)=cos acos b-sin asin b$.

$$
2cos 3x+cos 2x+1=0
$$

$$
2cos (2x+x)+cos 2x+1=0
$$

$$
2left(cos 2xcos x-sin 2xsin xright)+cos 2x+1=0
$$

$$
2cos 2xcos x-2sin 2xsin x+cos 2x+1=0
$$

Now we can use the double angle formulas of the sine and cosine functions where $color{#4257b2}sin 2x=2sin xcos x$ and $color{#4257b2}cos 2x=2cos^{2}x-1$.

$$
2left(2cos^{2}x-1right)cos x-2left(2sin xcos xright)sin x+left(2cos^{2}x-1right)+1=0
$$

$$
4cos^{3}x-2cos x-4sin^{2}xcos x+2cos^{2}x-1+1=0
$$

But we know from the Pythagorean identity that $color{#4257b2}sin^{2}x+cos^{2}x=1$, so $color{#4257b2}sin^{2}x=1-cos^{2}x$ and we can replace $color{#4257b2}sin^{2}x$ from our equation by $color{#4257b2}1-cos^{2}x$ to simplify.

$$
4cos^{3}x-2cos x-4left(1-cos^{2}xright)cos x+2cos^{2}x-1+1=0
$$

$$
4cos^{3}x-2cos x-4cos x+4cos^{3}x+2cos^{2}xcancel{-1}+cancel{1}=0
$$

$$
8cos^{3}x+2cos^{2}x-6cos x=0
$$

Now we note that all terms contain $color{#4257b2}2cos x$, so we can take it as a common factor.

$$
2cos xleft(4cos^{2}x+cos x-3right)=0
$$

Now we note that we have a quadratic trigonometric equation on the form $color{#4257b2}a y^{2} + b y+c=0$ where $color{#4257b2}y$ is considered to be $color{#4257b2}cos x$ in our equation, so we can factor to find the values of $color{#4257b2}cos x$.

$$
2cos xleft(cos x+1right)left(4cos x-3right)=0
$$

Now we can use the zero-factor property to find the values of $color{#4257b2}cos x$.

$$
2cos x=0, cos x+1=0 text{or} 4cos x-3=0
$$

$$
cos x=0, cos x=-1 text{or} cos x=dfrac{3}{4}
$$

Now we have three cases for $color{#4257b2}cos x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}cos x=0$

$$
cos^{-1}left(cos xright)=cos^{-1}left(0right)
$$

$$
x=cos^{-1}left(0right)
$$

$$
x=dfrac{pi}{2} text{or} x=dfrac{3pi}{2}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ and $color{#4257b2}dfrac{3pi}{2}$ are special angles which we know the value of the cosine function of them where $color{#4257b2}cos dfrac{pi}{2}=cos dfrac{3pi}{2}=0$.

So the solutions of the first case is $boxed{ x=dfrac{pi}{2} } text{or} boxed{ x=dfrac{3pi}{2} }$

For $color{#4257b2}cos x=-1$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-1right)
$$

$$
x=cos^{-1}left(-1right)
$$

$$
x=pi
$$

Note that $color{#4257b2}pi$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos pi=-1$.

So the solution of the second case is $boxed{ x=pi }$

For $color{#4257b2}cos x=dfrac{3}{4}$

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{3}{4}right)
$$

$$
x=cos^{-1}left(dfrac{3}{4}right)=0.72 text{radians}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos x=dfrac{3}{4}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos x=dfrac{3}{4}$ which means that it is positive, so the solutions in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
x=0.72 text{or} x=2pi-0.72=5.56 text{radians}
$$

So the solutions of the third case are $boxed{ x=0.72 text{radians} } text{or} boxed{ x=5.56 text{radians} }$

Now we found the solutions of each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{2}, pi, dfrac{3pi}{2}, 0.45 text{radians}, 5.56 text{radians}right}$

Large{$text{color{#c34632}$x=left{dfrac{pi}{2}, pi, dfrac{3pi}{2}, 0.45 text{radians}, 5.56 text{radians}right}$}$

We would like to solve the equation $color{#4257b2}3tan^{2}2x=1$ for $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. First, we can divide the two sides by $color{#4257b2}3$ to make $color{#4257b2}tan 2x$ in the left side alone.

$$
3tan^{2}2x=1
$$

$$
tan^{2}2x=dfrac{1}{3}
$$

Now we can take the square root for each side to find the values of $color{#4257b2}tan 2x$.

$$
tan 2x=pm sqrt{dfrac{1}{3}}
$$

$$
tan 2x=pm dfrac{1}{sqrt{3}}
$$

Now we have two cases for $color{#4257b2}tan 2x$, so we can solve each case to find the values of $color{#4257b2}x$.

For $color{#4257b2}tan 2x=dfrac{1}{sqrt{3}}$

$$
tan^{-1}left(tan 2xright)=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
2x=tan^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
2x=30text{textdegree}
$$

Note that $color{#4257b2}30text{textdegree}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan 30text{textdegree}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan 2x=dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan 2x=dfrac{1}{sqrt{3}}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $3$ and now we can use the related acute angle to find the solutions.

$$
2x=30text{textdegree} text{or} 2x=180text{textdegree}+30text{textdegree}
$$

$$
2x=30text{textdegree} text{or} 2x=210text{textdegree}
$$

$$
x=15text{textdegree} text{or} x=105text{textdegree}
$$

But we know that the period of $color{#4257b2}tan 2x$ is $color{#4257b2}dfrac{180text{textdegree}}{2}=90text{textdegree}$, so we can add this period two times only to the two solutions which we found in the final step to find all solutions in the interval $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. Note that we will add this period two times only because if we add it one more time the result will not be exist in the interval.

$$
x=15text{textdegree} text{or} x=105text{textdegree}
$$

$$
x=15text{textdegree}+90text{textdegree}=105text{textdegree} text{(already determined)} text{or} x=105text{textdegree}+90text{textdegree}=195text{textdegree}
$$

$$
x=105text{textdegree}+90text{textdegree}=195text{textdegree} text{(already determined)} text{or} x=195text{textdegree}+90text{textdegree}=285text{textdegree}
$$

So the solutions of the first case are $boxed{ x=15text{textdegree}, 105text{textdegree}, 195text{textdegree} } text{or} boxed{ x=285text{textdegree} }$

For $color{#4257b2}tan 2x=-dfrac{1}{sqrt{3}}$

$$
tan^{-1}left(tan 2xright)=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

$$
2x=tan^{-1}left(-dfrac{1}{sqrt{3}}right)
$$

Now we will calculate $color{#4257b2}tan^{-1}left(dfrac{1}{sqrt{3}}right)$ to find the related acute angle.

$$
2x=30text{textdegree}
$$

Note that $color{#4257b2}30text{textdegree}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan 30text{textdegree}=dfrac{1}{sqrt{3}}$.

Now we found the related acute angle for the equation $color{#4257b2}tan 2x=-dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}tan 2x=-dfrac{1}{sqrt{3}}$ which means that it is negative, so the solutions are in quadrant $2$ and quadrant $4$ and now we can use the related acute angle to find the solutions.

$$
2x=180text{textdegree}-30text{textdegree} text{or} 2x=360text{textdegree}-30text{textdegree}
$$

$$
2x=150text{textdegree} text{or} 2x=330text{textdegree}
$$

$$
x=75text{textdegree} text{or} x=165text{textdegree}
$$

But we know that the period of $color{#4257b2}tan 2x$ is $color{#4257b2}dfrac{180text{textdegree}}{2}=90text{textdegree}$, so we can add this period two times only to the two solutions which we found in the final step to find all solutions in the interval $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$. Note that we will add this period two times only because if we add it one more time the result will not be exist in the interval.

$$
x=75text{textdegree} text{or} x=165text{textdegree}
$$

$$
x=75text{textdegree}+90text{textdegree}=165text{textdegree} text{(already determined)} text{or} x=165text{textdegree}+90text{textdegree}=255text{textdegree}
$$

$$
x=165text{textdegree}+90text{textdegree}=255text{textdegree} text{(already determined)} text{or} x=255text{textdegree}+90text{textdegree}=345text{textdegree}
$$

So the solutions of the second case are $boxed{ x=75text{textdegree}, 165text{textdegree}, 255text{textdegree} } text{or} boxed{ x=345text{textdegree} }$

Now we found the solutions for each case, so the solutions of the equation are $color{#4257b2}x=left{15text{textdegree}, 75text{textdegree}, 105text{textdegree}, 165text{textdegree}, 195text{textdegree}, 255text{textdegree}, 285text{textdegree}, 345text{textdegree}right}$

Large{$text{color{#c34632}$x=left{15text{textdegree}, 75text{textdegree}, 105text{textdegree}, 165text{textdegree}, 195text{textdegree}, 255text{textdegree}, 285text{textdegree}, 345text{textdegree}right}$}$

We would like to solve the equation $color{#4257b2}sqrt{2} sin theta=sqrt{3} – cos theta$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can square the two sides to simplify.

$$
sqrt{2} sin theta=sqrt{3} – cos theta
$$

$$
left(sqrt{2} sin thetaright)^{2}=left(sqrt{3} – cos thetaright)^{2}
$$

$$
2sin^{2}theta=3-2sqrt{3} cos theta+cos^{2}theta
$$

Note that $color{#4257b2}(a+b)^{2}=a^{2}+2ab+b^{2}$. Now since we know from the Pythagorean identity that $color{#4257b2}sin^{2}theta+cos^{2}theta=1$, so $color{#4257b2}sin^{2}theta=1-cos^{2}theta$ and we can use this identity to replace $color{#4257b2}sin^{2}theta$ from the left side by $color{#4257b2}1-cos^{2}theta$.

$$
2sin^{2}theta=3-2sqrt{3} cos theta+cos^{2}theta
$$

$$
2left(1-cos^{2}thetaright)=3-2sqrt{3} cos theta+cos^{2}theta
$$

$$
2-2cos^{2}theta=3-2sqrt{3} cos theta+cos^{2}theta
$$

Now we can add $color{#4257b2}2cos^{2}theta-2$ to each side to make the left side equals zero.

$$
2-2cos^{2}theta+2cos^{2}theta-2=3-2sqrt{3} cos theta+cos^{2}theta+2cos^{2}theta-2
$$

$$
0=1-2sqrt{3} cos theta+3cos^{2}theta
$$

$$
3cos^{2}theta-2sqrt{3} cos theta+1=0
$$

Now we have a quadratic equation on the form $color{#4257b2}a x^{2} + b x+c=0$ where $color{#4257b2}x$ is considered to be $color{#4257b2}cos theta$ in our equation, so we can factor to find the values of $color{#4257b2}cos theta$

$$
3cos^{2}theta-2sqrt{3} cos theta+1=0
$$

$$
left(sqrt{3} cos theta-1right)left(sqrt{3} cos theta-1right)=0
$$

Now we can use the zero-factor property.

$$
sqrt{3} cos theta-1=0 text{or} sqrt{3} cos theta-1=0
$$

$$
cos theta=dfrac{1}{sqrt{3}} text{or} cos theta=dfrac{1}{sqrt{3}}
$$

Now we have two similar equations for $color{#4257b2}cos theta$, so we can solve one of them to find the values of $color{#4257b2}theta$.

$$
cos theta=dfrac{1}{sqrt{3}}
$$

We can take $color{#4257b2}cos^{-1}$ for each side to find the related acute angle.

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
theta=cos^{-1}left(dfrac{1}{sqrt{3}}right)
$$

$$
theta=0.96 text{radian}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos theta=dfrac{1}{sqrt{3}}$, so the next step is to know in which quadrants the solutions are exist. We note that $color{#4257b2}cos theta=dfrac{1}{sqrt{3}}$ which means that it is positive, so the solutions are in quadrant $1$ and quadrant $4$ where the cosine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions.

$$
theta=0.96 text{or} theta=2pi-0.96
$$

$$
theta=0.96 text{radians} text{or} theta=5.33 text{radians}
$$

Now we found two values for $color{#4257b2}theta$, so the next step is to check these values by substituting them in the original equation and find if they satisfy the original equation or not. Note that we will do this check because we squared the equation from the beginning, so there may be solutions which doesn’t satisfy the original equation.

For the solution $color{#4257b2}theta=0.96 text{radians}$

$$
sqrt{2} sin theta=sqrt{3}-cos theta
$$

$$
sqrt{2} sin left(0.96 text{radians}right)=sqrt{3}-cos left(0.96 text{radians}right)
$$

$$
1.1547=sqrt{3}-0.57735
$$

$$
1.1547=1.1547
$$

So the solution $color{#4257b2}theta=0.96 text{radians}$ is true because it satisfies the original equation.

For the solution $color{#4257b2}theta=5.33 text{radians}$

$$
sqrt{2} sin theta=sqrt{3}-cos theta
$$

$$
sqrt{2} sin left(5.33 text{radians}right)=sqrt{3}-cos left(5.33 text{radians}right)
$$

$$
-1.1547=sqrt{3}-0.57735
$$

$$
-1.1547=1.1547
$$

So the solution $color{#4257b2}theta=5.33 text{radians}$ is false because it doesn’t satisfy the original equation.

So the solution of the equation is $boxed{ theta=0.96 text{radians} }$

$$
color{#c34632}theta=0.96 text{radians}
$$