(a) We would like to use the graph of $color{#4257b2}y=sin theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}sin theta=1
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}sin theta=1$ since we have the graph of $color{#4257b2}y=sin theta$, so we can graph the line $color{#4257b2}y=1$ on the same plane of the graph of $color{#4257b2}y=sin theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the point $color{#4257b2}A$ and the value of $color{#4257b2}theta$ at this point is $color{#4257b2}dfrac{pi}{2}$, so the solution of the equation $color{#4257b2}sin theta=1$ is $boxed{ theta=dfrac{pi}{2} }$

(b) We would like to use the graph of $color{#4257b2}y=sin theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}sin theta=-1
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}sin theta=-1$ since we have the graph of $color{#4257b2}y=sin theta$, so we can graph the line $color{#4257b2}y=-1$ on the same plane of the graph of $color{#4257b2}y=sin theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the point $color{#4257b2}A$ and the value of $color{#4257b2}theta$ at this point is $color{#4257b2}dfrac{3pi}{2}$, so the solution of the equation $color{#4257b2}sin theta=-1$ is $boxed{ theta=dfrac{3pi}{2} }$

(c) We would like to use the graph of $color{#4257b2}y=sin theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}sin theta=0.5
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}sin theta=0.5$ since we have the graph of $color{#4257b2}y=sin theta$, so we can graph the line $color{#4257b2}y=0.5$ on the same plane of the graph of $color{#4257b2}y=sin theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{pi}{6}$ and $color{#4257b2}dfrac{5pi}{6}$, so the solutions of the equation $color{#4257b2}sin theta=0.5$ are $boxed{ theta=dfrac{pi}{6} text{or} theta=dfrac{5pi}{6} }$

(d) We would like to use the graph of $color{#4257b2}y=sin theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}sin theta=-0.5
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}sin theta=-0.5$ since we have the graph of $color{#4257b2}y=sin theta$, so we can graph the line $color{#4257b2}y=-0.5$ on the same plane of the graph of $color{#4257b2}y=sin theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{7pi}{6}$ and $color{#4257b2}dfrac{11pi}{6}$, so the solutions of the equation $color{#4257b2}sin theta=-0.5$ are $boxed{ theta=dfrac{7pi}{6} text{or} theta=dfrac{11pi}{6} }$

(e) We would like to use the graph of $color{#4257b2}y=sin theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}sin theta=0
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}sin theta=0$ since we have the graph of $color{#4257b2}y=sin theta$, so we need to find the points which the graph of $color{#4257b2}y=sin theta$ will intersect with the x-axis and then find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the graph $color{#4257b2}y=sin theta$ intersects with the x-axis at the points $color{#4257b2}A, B$ and $color{#4257b2}C$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}0, pi$ and $color{#4257b2}2pi$, so the solutions of the equation $color{#4257b2}sin theta=0$ are $boxed{ theta=0, pi text{and} theta=2pi }$

(f) We would like to use the graph of $color{#4257b2}y=sin theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}sin theta=dfrac{sqrt{3}}{2}
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}sin theta=dfrac{sqrt{3}}{2}$ since we have the graph of $color{#4257b2}y=sin theta$, so we can graph the line $color{#4257b2}y=dfrac{sqrt{3}}{2}$ on the same plane of the graph of $color{#4257b2}y=sin theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{pi}{3}$ and $color{#4257b2}dfrac{2pi}{3}$, so the solutions of the equation $color{#4257b2}sin theta=dfrac{sqrt{3}}{2}$ are $boxed{ theta=dfrac{pi}{3} text{or} theta=dfrac{2pi}{3} }$

$$
text{color{#c34632}$(a) theta=dfrac{pi}{2}$ $(c) theta=dfrac{pi}{6} {color{Black}text{or}} theta=dfrac{5pi}{6}$ $(e) theta=0, pi {color{Black}text{and}} theta=2pi$
\
\
\
color{#c34632}$(b) theta=dfrac{3pi}{2}$ $(d) theta=dfrac{7pi}{6} {color{Black}text{or}} theta=dfrac{11pi}{6}$ $(f) theta=dfrac{pi}{3} {color{Black}text{or}} theta=dfrac{2pi}{3}$}
$$

(a) We would like to use the graph of $color{#4257b2}y=cos theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}cos theta=1
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}cos theta=1$ since we have the graph of $color{#4257b2}y=cos theta$, so we can graph the line $color{#4257b2}y=1$ on the same plane of the graph of $color{#4257b2}y=cos theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}0$ and $color{#4257b2}2pi$, so the solutions of the equation $color{#4257b2}cos theta=1$ are $boxed{ theta=0 } text{or} boxed{ theta=2pi }$

(b) We would like to use the graph of $color{#4257b2}y=cos theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}cos theta=-1
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}cos theta=-1$ since we have the graph of $color{#4257b2}y=cos theta$, so we can graph the line $color{#4257b2}y=-1$ on the same plane of the graph of $color{#4257b2}y=cos theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the point $color{#4257b2}A$ and the value of $color{#4257b2}theta$ at this point is $color{#4257b2}pi$, so the solution of the equation $color{#4257b2}cos theta=-1$ is $boxed{ theta=pi }$

(c) We would like to use the graph of $color{#4257b2}y=cos theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}cos theta=0.5
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}cos theta=0.5$ since we have the graph of $color{#4257b2}y=cos theta$, so we can graph the line $color{#4257b2}y=0.5$ on the same plane of the graph of $color{#4257b2}y=cos theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{pi}{3}$ and $color{#4257b2}dfrac{5pi}{3}$, so the solutions of the equation $color{#4257b2}cos theta=0.5$ are $boxed{ theta=dfrac{pi}{3} text{or} theta=dfrac{5pi}{3} }$

(d) We would like to use the graph of $color{#4257b2}y=cos theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}cos theta=-0.5
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}cos theta=-0.5$ since we have the graph of $color{#4257b2}y=cos theta$, so we can graph the line $color{#4257b2}y=-0.5$ on the same plane of the graph of $color{#4257b2}y=cos theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{2pi}{3}$ and $color{#4257b2}dfrac{4pi}{3}$, so the solutions of the equation $color{#4257b2}cos theta=-0.5$ are $boxed{ theta=dfrac{2pi}{3} text{or} theta=dfrac{4pi}{3} }$

(e) We would like to use the graph of $color{#4257b2}y=cos theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}cos theta=0
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}cos theta=0$ since we have the graph of $color{#4257b2}y=cos theta$, so we need to find the points which the graph of $color{#4257b2}y=cos theta$ will intersect with the x-axis and then find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the graph $color{#4257b2}y=cos theta$ intersects with the x-axis at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{pi}{2}$ and $color{#4257b2}dfrac{3pi}{2}$, so the solutions of the equation $color{#4257b2}cos theta=0$ are $boxed{ theta=dfrac{pi}{2} text{or} theta=dfrac{3pi}{2} }$

(f) We would like to use the graph of $color{#4257b2}y=cos theta$ to estimate the values of $color{#4257b2}theta$ in the interval $color{#4257b2}0 leq theta leq 2pi$ for the equation
$$
color{#4257b2}cos theta=dfrac{sqrt{3}}{2}
$$

First, we need to find the values of $color{#4257b2}theta$ for the equation $color{#4257b2}cos theta=dfrac{sqrt{3}}{2}$ since we have the graph of $color{#4257b2}y=cos theta$, so we can graph the line $color{#4257b2}y=dfrac{sqrt{3}}{2}$ on the same plane of the graph of $color{#4257b2}y=cos theta$ and find the points which the two graphs will intersect and find the values of $color{#4257b2}theta$ at these points.

We note from the graphs that the two graphs intersect at the points $color{#4257b2}A$ and $color{#4257b2}B$ and the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{pi}{6}$ and $color{#4257b2}dfrac{11pi}{6}$, so the solutions of the equation $color{#4257b2}cos theta=dfrac{sqrt{3}}{2}$ are $boxed{ theta=dfrac{pi}{6} text{or} theta=dfrac{11pi}{6} }$

$$
text{color{#c34632}$(a) 0 {color{Black}text{or}} theta=2pi$ $(c) theta=dfrac{pi}{3} {color{Black}text{or}} theta=dfrac{5pi}{3}$ $(e) theta=dfrac{pi}{2} {color{Black}text{or}} theta=dfrac{3pi}{2}$
\
\
\
color{#c34632}$(b) theta=pi$ $(d) theta=dfrac{2pi}{3} {color{Black}text{or}} theta=dfrac{4pi}{3}$ $(f) theta=dfrac{pi}{6} {color{Black}text{or}} theta=dfrac{11pi}{6}$}
$$

We would like to solve the equation $color{#4257b2}sin x=dfrac{sqrt{3}}{2}$, where $color{#4257b2}0 leq x leq 2pi$.

(a) We would like to know how many solutions are possible for our equation. Since the sine ratio is positive and we know that $color{#4257b2}sin theta=dfrac{text{opposite}}{text{hypotenuse}}$, so we need the opposite to be positive and this may happen in two cases if the opposite and adjacent are positive or if the opposite is positive and the adjacent is negative, so there are two possible solutions for the equation $color{#4257b2}sin x=dfrac{sqrt{3}}{2}$.

(b) We would like to know in which quadrants can we find the solutions. Since we know that $color{#4257b2}sin x$ is positive, so the solutions are in quadrant $1$ and quadrant $2$ because we know that the sine ratio is positive in quadrant $1$ and quadrant $2$.

(c) We would like to determine the related acute angle for the equation. First, we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

$$
sin x=dfrac{sqrt{3}}{2}
$$

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$.

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

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

So the related acute angle for the equation $color{#4257b2}sin x=dfrac{sqrt{3}}{2}$ is $boxed{ x=dfrac{pi}{3} }$

(d) Since we would like to determine all solutions for the equation, we can use the related acute angle $color{#4257b2}dfrac{pi}{3}$ to find the solutions where we know that these solutions will be in quadrant $1$ and quadrant $2$.

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

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

So the solutions for our equation in the interval $color{#4257b2}0 leq x leq 2pi$ are $boxed{ x=dfrac{pi}{3} } text{or} boxed{ x=dfrac{5pi}{3} }$

(a) $text{color{Black} There are {color{#c34632}two} possible solutions}$

(b) $text{color{#c34632} {color{Black}The two solutions are in} quadrant $1$ {color{Black}and} quadrant $2$}$

(c) $text{color{#c34632} {color{Black}The related acute angle is} $x=dfrac{pi}{3}$}$

(d) $text{color{#c34632} {color{Black}The solutions are} $x=dfrac{pi}{3}$ {color{Black}text{or}} $x=dfrac{5pi}{3}$}$

We would like to solve the equation $color{#4257b2}cos x=-0.8667$, where $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$.

(a) We would like to know how many solutions are possible for our equation. Since the cosine ratio is negative and we know that $color{#4257b2}cos theta=dfrac{text{adjacent}}{text{hypotenuse}}$, so we need the adjacent to be negative and this may happen in two cases if the adjacent and opposite are negative or if the adjacent is negative and the opposite is positive, so there are two possible solutions for the equation $color{#4257b2}cos x=-0.8667$.

(b) We would like to know in which quadrants can we find the solutions. Since we know that $color{#4257b2}cos x$ is negative, so the solutions are in quadrant $2$ and quadrant $3$ because we know that the cosine ratio is negative in quadrant $2$ and quadrant $3$.

(c) We would like to determine the related acute angle for the equation. First, we can take $color{#4257b2}cos^{-1}$ for each side to find the value of the related acute angle.

$$
cos x=-0.8667
$$

$$
cos^{-1}left(cos xright)=cos^{-1}left(-0.8667right)
$$

$$
x=cos^{-1}left(-0.8667right)
$$

Note that $color{#4257b2}cos^{-1}left(cos thetaright)=theta$. Now we can use the calculator and write $color{#4257b2}cos^{-1}left(0.8667right)$ to find the related acute angle. Note that we will not write $color{#4257b2}cos^{-1}left(-0.8667right)$ because we need to find the acute angle.

$$
x=cos^{-1}left(-0.8667right)
$$

$$
x=29.9text{textdegree}
$$

So the related acute angle for the equation $color{#4257b2}cos x=-0.8667$ is $boxed{ x=29.9text{textdegree} }$

(d) Since we would like to determine all solutions for the equation, we can use the related acute angle $color{#4257b2}29.9text{textdegree}$ to find the solutions where we know that these solutions will be in quadrant $2$ and quadrant $3$.

$$
x=180text{textdegree}-29.9text{textdegree} text{or} x=180text{textdegree}+29.9text{textdegree}
$$

$$
x=150.1text{textdegree} text{or} x=209.9text{textdegree}
$$

So the solutions for our equation in the interval $color{#4257b2}0text{textdegree} leq x leq 360text{textdegree}$ are $boxed{ x=150.1text{textdegree} } text{or} boxed{ x=209.9text{textdegree} }$

(a) $text{color{Black} There are {color{#c34632}two} possible solutions}$

(b) $text{color{#c34632} {color{Black}The two solutions are in} quadrant $2$ {color{Black}and} quadrant $3$}$

(c) $text{color{#c34632} {color{Black}The related acute angle is} $x=29.9text{textdegree}$}$

(d) $text{color{#c34632} {color{Black}The solutions are} $x=150.1text{textdegree}$ {color{Black}text{or}} $x=209.9text{textdegree}$}$

We would like to solve the equation $color{#4257b2}tan x=2.7553$, where $color{#4257b2}0 leq x leq 2pi$.

(a) We would like to know how many solutions are possible for our equation. Since the tangent ratio is positive and we know that $color{#4257b2}tan theta=dfrac{text{opposite}}{text{adjacent}}$, so we need the tangent ratio to be positive and this may happen in two cases if the opposite and adjacent are positive or if the opposite and adjacent are negative, so there are two possible solutions for the equation $color{#4257b2}tan x=2.7553$.

(b) We would like to know in which quadrants can we find the solutions. Since we know that $color{#4257b2}tan x$ is positive, so the solutions are in quadrant $1$ and quadrant $3$ because we know that the tangent ratio is positive in quadrant $1$ and quadrant $3$.

(c) We would like to determine the related acute angle for the equation. First, we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

$$
tan x=2.7553
$$

$$
tan^{-1}left(tan xright)=tan^{-1}left(2.7553right)
$$

$$
x=tan^{-1}left(2.7553right)
$$

Note that $color{#4257b2}tan^{-1}left(tan thetaright)=theta$. Now we can use the calculator and write $color{#4257b2}tan^{-1}left(2.7553right)$ to find the acute angle.

$$
x=tan^{-1}left(2.7553right)
$$

$$
x=1.22 text{radians}
$$

So the related acute angle for the equation $color{#4257b2}tan x=2.7553$ is $boxed{ x=1.22 text{radians} }$

(d) Since we would like to determine all solutions for the equation, we can use the related acute angle $color{#4257b2}1.22 text{radians}$ to find the solutions where we know that these solutions will be in quadrant $1$ and quadrant $3$.

$$
x=1.22 text{or} x=pi+1.22
$$

$$
x=1.22 text{radians} text{or} x=4.36 text{radians}
$$

So the solutions for our equation in the interval $color{#4257b2}0 leq x leq 2pi$ are
$$
boxed{ x=1.22 text{radians} } text{or} boxed{ x=4.36 text{radians} }
$$

(a) $text{color{Black} There are {color{#c34632}two} possible solutions}$

(b) $text{color{#c34632} {color{Black}The two solutions are in} quadrant $1$ {color{Black}and} quadrant $3$}$

(c) $text{color{#c34632} {color{Black}The related acute angle is} $x=1.22 text{radians}$}$

(d) $text{color{#c34632} {color{Black}The solutions are} $x=1.22 text{radians}$ {color{Black}text{or}} $x=4.36 text{radians}$}$

(a) We would like to determine the solutions for the equation $color{#4257b2}tan theta=1$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

$$
tan theta=1
$$

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

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

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

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

Note that $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the tangent function for 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 of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}tan theta=1$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ because we know that the tangent ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq theta leq 2pi$.

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

$$
theta=dfrac{pi}{4} text{or} theta=dfrac{5pi}{4}
$$

So the solutions of the equation are $boxed{ theta=dfrac{pi}{4} } text{or} boxed{ theta=dfrac{5pi}{4} }$

(b) We would like to determine the solutions for the equation $color{#4257b2}sin theta=dfrac{1}{sqrt{2}}$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$.

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

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

Now we found the related acute angle for the equation $color{#4257b2}sin theta=dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin theta=dfrac{1}{sqrt{2}}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $2$ because we know that the sine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq theta leq 2pi$.

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

$$
theta=dfrac{pi}{4} text{or} theta=dfrac{3pi}{4}
$$

So the solutions of the equation are $boxed{ theta=dfrac{pi}{4} } text{or} boxed{ theta=dfrac{3pi}{4} }$

(c) We would like to determine the solutions for the equation $color{#4257b2}cos theta=dfrac{sqrt{3}}{2}$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can take $color{#4257b2}cos^{-1}$ for each side to find the value of the related acute angle.

$$
cos theta=dfrac{sqrt{3}}{2}
$$

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

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

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

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

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

Now we found the related acute angle for the equation $color{#4257b2}cos theta=dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos theta=dfrac{sqrt{3}}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $4$ because we know that the cosine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq theta leq 2pi$.

$$
theta=dfrac{pi}{6} text{or} theta=2pi-dfrac{pi}{6}
$$

$$
theta=dfrac{pi}{6} text{or} theta=dfrac{11pi}{6}
$$

So the solutions of the equation are $boxed{ theta=dfrac{pi}{6} } text{or} boxed{ theta=dfrac{11pi}{6} }$

(d) We would like to determine the solutions for the equation $color{#4257b2}sin theta=-dfrac{sqrt{3}}{2}$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

$$
sin theta=-dfrac{sqrt{3}}{2}
$$

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{sqrt{3}}{2}right)$ not $color{#4257b2}sin^{-1}left(-dfrac{sqrt{3}}{2}right)$ to find the acute angle.

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

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

Now we found the related acute angle for the equation $color{#4257b2}sin theta=-dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin theta=-dfrac{sqrt{3}}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ because we know that the sine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq theta leq 2pi$.

$$
theta=pi+dfrac{pi}{3}=dfrac{4pi}{3} text{or} theta=2pi-dfrac{pi}{3}=dfrac{5pi}{3}
$$

So the solutions of the equation are $boxed{ theta=dfrac{4pi}{3} } text{or} boxed{ theta=dfrac{5pi}{3} }$

(e) We would like to determine the solutions for the equation $color{#4257b2}cos theta=-dfrac{1}{sqrt{2}}$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can take $color{#4257b2}cos^{-1}$ for each side to find the value of the related acute angle.

$$
cos theta=-dfrac{1}{sqrt{2}}
$$

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

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

Note that $color{#4257b2}cos^{-1}left(cos thetaright)=theta$. Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{sqrt{2}}right)$ not $color{#4257b2}cos^{-1}left(-dfrac{1}{sqrt{2}}right)$ to find the acute angle.

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

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

Now we found the related acute angle for the equation $color{#4257b2}cos theta=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos theta=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions will be in quadrant $2$ and quadrant $3$ because we know that the cosine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq theta leq 2pi$.

$$
theta=pi-dfrac{pi}{4}=dfrac{3pi}{4} text{or} theta=pi+dfrac{pi}{4}=dfrac{5pi}{4}
$$

So the solutions of the equation are $boxed{ theta=dfrac{3pi}{4} } text{or} boxed{ theta=dfrac{5pi}{4} }$

(f) We would like to determine the solutions for the equation $color{#4257b2}tan theta=sqrt{3}$, where $color{#4257b2}0 leq theta leq 2pi$. First, we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

$$
tan theta=sqrt{3}
$$

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

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

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

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

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

Now we found the related acute angle for the equation $color{#4257b2}tan theta=sqrt{3}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}tan theta=1$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ because we know that the tangent ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq theta leq 2pi$.

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

$$
theta=dfrac{pi}{3} text{or} theta=dfrac{4pi}{3}
$$

So the solutions of the equation are $boxed{ theta=dfrac{pi}{3} } text{or} boxed{ theta=dfrac{4pi}{3} }$

Large{$text{$text{$text{color{#c34632}(a) $theta=dfrac{pi}{4} {color{Black}text{or}} theta=dfrac{5pi}{4}$ (b) $theta=dfrac{pi}{4} {color{Black}text{or}} theta=dfrac{3pi}{4}$
\
\
\
Large{color{#c34632}(c) $theta=dfrac{pi}{6} {color{Black}text{or}} theta=dfrac{11pi}{6}$ (d) $theta=dfrac{4pi}{3} {color{Black}text{or}} theta=dfrac{5pi}{3}$
\
\
\
Large{color{#c34632}(e) $theta=dfrac{3pi}{4} {color{Black}text{or}} theta=dfrac{5pi}{4}$ (f) $theta=dfrac{pi}{3} {color{Black}text{or}} theta=dfrac{4pi}{3}$}$}$}$

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

$$
2sin theta=-1
$$

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

Now we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{2}right)$ not $color{#4257b2}sin^{-1}left(-dfrac{1}{2}right)$ to find the acute angle.

$$
theta=30text{textdegree}
$$

Note that $color{#4257b2}30text{textdegree}$ is a special angle which we know the value of the sine function for it where $color{#4257b2}sin 30text{textdegree}=dfrac{1}{2}$.

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 of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin theta=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ because we know that the sine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$.

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

$$
theta=210text{textdegree} text{or} theta=330text{textdegree}
$$

So the solutions of the equation are $boxed{ theta=210text{textdegree} } text{or} boxed{ theta=330text{textdegree} }$

(b) We would like to determine the solutions for the equation $color{#4257b2}3cos theta=-2$, where $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we can divide the two sides by $3$ to make $color{#4257b2}cos theta$ in the left side alone.

$$
3cos theta=-2
$$

$$
cos theta=-dfrac{2}{3}
$$

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

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

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

Note that $color{#4257b2}cos^{-1}left(cos thetaright)=theta$. Now we will use the calculator to determine $color{#4257b2}cos^{-1}left(dfrac{2}{3}right)$ not $color{#4257b2}cos^{-1}left(-dfrac{2}{3}right)$ to find the acute angle.

$$
theta=48.2text{textdegree}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos theta=-dfrac{2}{3}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos theta=-dfrac{2}{3}$ which means that it is negative, so the solutions will be in quadrant $2$ and quadrant $3$ because we know that the cosine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$.

$$
theta=180text{textdegree}-48.2text{textdegree} text{or} theta=180text{textdegree}+48.2text{textdegree}
$$

$$
theta=131.8text{textdegree} text{or} theta=228.2text{textdegree}
$$

So the solutions of the equation are $boxed{ theta=131.8text{textdegree} } text{or} boxed{ theta=228.2text{textdegree} }$

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

$$
2tan theta=3
$$

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

Now we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}tan^{-1}left(tan thetaright)=theta$. Now we will use the calculator to determine $color{#4257b2}tan^{-1}left(dfrac{3}{2}right)$ to find the acute angle.

$$
theta=56.3text{textdegree}
$$

Now we found the related acute angle for the equation $color{#4257b2}tan theta=dfrac{3}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}tan theta=dfrac{3}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ because we know that the tangent ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$.

$$
theta=56.3text{textdegree} text{or} theta=180text{textdegree}+56.3text{textdegree}
$$

$$
theta=56.3text{textdegree} text{or} theta=236.3text{textdegree}
$$

So the solutions of the equation are $boxed{ theta=56.3text{textdegree} } text{or} boxed{ theta=236.3text{textdegree} }$

(d) We would like to determine the solutions for the equation $color{#4257b2}-3sin theta-1=1$, where $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we will make $color{#4257b2}sin theta$ in the left side alone.

$$
-3sin theta-1=1
$$

$$
-3sin theta=1+1=2
$$

$$
sin theta=-dfrac{2}{3}
$$

Now we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$. Now we will use the calculator to determine $color{#4257b2}sin^{-1}left(dfrac{2}{3}right)$ not $color{#4257b2}sin^{-1}left(-dfrac{2}{3}right)$ to find the acute angle.

$$
theta=41.8text{textdegree}
$$

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 of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin theta=-dfrac{2}{3}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ because we know that the sine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$.

$$
theta=180text{textdegree}+41.8text{textdegree}=221.8text{textdegree} text{or} theta=360text{textdegree}-41.8text{textdegree}=318.2text{textdegree}
$$

So the solutions of the equation are $boxed{ theta=221.8text{textdegree} } text{or} boxed{ theta=318.2text{textdegree} }$

(e) We would like to determine the solutions for the equation $color{#4257b2}-5cos theta+3=2$, where $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we will make $color{#4257b2}cos theta$ in the left side alone.

$$
-5cos theta+3=2
$$

$$
-5cos theta=2-3=-1
$$

$$
cos theta=dfrac{-1}{-5}=dfrac{1}{5}
$$

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

$$
cos^{-1}left(cos thetaright)=cos^{-1}left(dfrac{1}{5}right)
$$

$$
theta=cos^{-1}left(dfrac{1}{5}right)
$$

Note that $color{#4257b2}cos^{-1}left(tan thetaright)=theta$. Now we will use the calculator to determine $color{#4257b2}cos^{-1}left(dfrac{1}{5}right)$ to find the acute angle.

$$
theta=78.5text{textdegree}
$$

Now we found the related acute angle for the equation $color{#4257b2}cos theta=dfrac{1}{5}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos theta=dfrac{1}{5}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $4$ because we know that the cosine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$.

$$
theta=78.5text{textdegree} text{or} theta=360text{textdegree}-78.5text{textdegree}=281.5text{textdegree}
$$

So the solutions of the equation are $boxed{ theta=78.5text{textdegree} } text{or} boxed{ theta=281.5text{textdegree} }$

(f) We would like to determine the solutions for the equation $color{#4257b2}8-tan theta=10$, where $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$. First, we will make $color{#4257b2}tan theta$ in the left side alone.

$$
8-tan theta=10
$$

$$
-tan theta=10-8=2
$$

$$
tan theta=dfrac{2}{-1}=-2
$$

Now we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}tan^{-1}left(tan thetaright)=theta$. Now we will use the calculator to determine $color{#4257b2}tan^{-1}left(2right)$ not $color{#4257b2}tan^{-1}left(-2right)$ to find the acute angle.

$$
theta=63.4text{textdegree}
$$

Now we found the related acute angle for the equation $color{#4257b2}tan theta=-2$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}tan theta=-2$ which means that it is negative, so the solutions will be in quadrant $2$ and quadrant $4$ because we know that the tangent ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0text{textdegree} leq theta leq 360text{textdegree}$.

$$
theta=180text{textdegree}-63.4text{textdegree}=116.6text{textdegree} text{or} theta=360text{textdegree}-63.4text{textdegree}=296.6text{textdegree}
$$

So the solutions of the equation are $boxed{ theta=116.6text{textdegree} } text{or} boxed{ theta=296.6text{textdegree} }$

Large{$text{$text{$text{color{#c34632}(a) $theta=210text{textdegree} {color{Black}text{or}} theta=330text{textdegree}$ (b) $theta=131.8text{textdegree} {color{Black}text{or}} theta=228.2text{textdegree}$
\
\
\
Large{color{#c34632}(c) $theta=56.3text{textdegree} {color{Black}text{or}} theta=236.3text{textdegree}$ (d) $theta=221.8text{textdegree} {color{Black}text{or}} theta=318.2text{textdegree}$
\
\
\
Large{color{#c34632}(e) $theta=78.5text{textdegree} {color{Black}text{or}} theta=281.5text{textdegree}$ (f) $theta=116.6text{textdegree} {color{Black}text{or}} theta=296.6text{textdegree}$}$}$}$

(a) We would like to determine the solutions for the equation
$color{#4257b2}3sin x=sin x+1$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}sin x$ in the left side alone.

$$
3sin x=sin x+1
$$

$$
2sin x=1
$$

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

Now we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}sin^{-1}left(sin xright)=x$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{2}right)$ 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 sine function for 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 our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin x=dfrac{1}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $2$ because we know that the sine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

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

So the solutions of the equation are $boxed{ x=0.52 } text{or} boxed{ x=2.62 }$

(b) We would like to determine the solutions for the equation
$color{#4257b2}5cos x-sqrt{3}=3cos x$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}cos x$ in the left side alone.

$$
5cos x-sqrt{3}=3cos x
$$

$$
5cos x-sqrt{3}-3cos x+sqrt{3}=3cos x-3cos x+sqrt{3}
$$

$$
5cos xcancel{-sqrt{3}}-3cos x+cancel{sqrt{3}}=cancel{3cos x}cancel{-3cos x}+sqrt{3}
$$

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

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

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

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

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

Note that $color{#4257b2}cos^{-1}left(cos xright)=x$. Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{sqrt{3}}{2}right)$ 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 cosine function for 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 of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos x=dfrac{sqrt{3}}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $4$ because we know that the cosine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$$
x=dfrac{pi}{6}=0.52 text{or} x=dfrac{11pi}{6}=5.76
$$

So the solutions of the equation are $boxed{ x=0.52 } text{or} boxed{ x=5.76 }$

(c) We would like to determine the solutions for the equation
$color{#4257b2}cos x-1=-cos x$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}cos x$ in the left side alone.

$$
cos x-1=-cos x
$$

$$
cos x-1+cos x+1=-cos x+cos x+1
$$

$$
cos xcancel{-1}+cos x+cancel{1}=cancel{-cos x}+cancel{cos x}+1
$$

$$
2cos x=1
$$

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

Now we can take $color{#4257b2}cos^{-1}$ for each side to find the value of 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 $color{#4257b2}cos^{-1}left(cos xright)=x$. Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ to find the 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 for 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 our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos x=dfrac{1}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $4$ because we know that the cosine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

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

So the solutions of the equation are $boxed{ x=1.05 } text{or} boxed{ x=5.24 }$

(d) We would like to determine the solutions for the equation
$color{#4257b2}5sin x+1=3sin x$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}sin x$ in the left side alone.

$$
5sin x+1=3sin x
$$

$$
5sin x+1-3sin x-1=3sin x-3sin x-1
$$

$$
5sin x+cancel{1}-3sin xcancel{-1}=cancel{3sin x}cancel{-3sin x}-1
$$

$$
2sin x=-1
$$

$$
sin x=-dfrac{1}{2}
$$

Now we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}sin^{-1}left(sin xright)=x$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{2}right)$ not $color{#4257b2}sin^{-1}left(-dfrac{1}{2}right)$ 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 sine function for 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 our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ because we know that the sine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$$
x=dfrac{7pi}{6}=3.67 text{or} x=dfrac{11pi}{6}=5.76
$$

So the solutions of the equation are $boxed{ x=3.67 } text{or} boxed{ x=5.76 }$

Large{$text{$text{color{#c34632}(a) $x=0.52 text{radians} {color{Black}text{or}} x=2.62 text{radians}$
\
\
(b) $x=0.52 text{radians} {color{Black}text{or}} x=5.76 text{radians}$
\
\
Large{color{#c34632}(c) $x=1.05 text{radians} {color{Black}text{or}} x=5.24 text{radians}$
\
\
(d) $x=3.67 text{radians} {color{Black}text{or}} x=5.76 text{radians}$}$}$

(a) We would like to determine the solutions for the equation
$color{#4257b2}2-2cot x=0$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}cot x$ in the left side alone.

$$
2-2cot x=0
$$

$$
-2cot x=-2
$$

$$
cot x=dfrac{-2}{-2}=1
$$

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

$$
dfrac{1}{cot x}=dfrac{1}{1}
$$

$$
tan x=1
$$

Now we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}tan^{-1}left(tan xright)=x$. Now we will calculate $color{#4257b2}tan^{-1}left(1right)$ to find the 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 for 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 of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}tan x=1$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ because we know that the tangent ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

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

So the solutions of the equation are $boxed{ x=0.79 } text{or} boxed{ x=3.93 }$

(b) We would like to determine the solutions for the equation
$color{#4257b2}csc x-2=0$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}csc x$ in the left side alone.

$$
csc x-2=0
$$

$$
csc x=2
$$

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

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

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

Now we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}sin^{-1}left(sin xright)=x$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{2}right)$ 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 sine function for 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 our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin x=dfrac{1}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $2$ because we know that the sine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

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

So the solutions of the equation are $boxed{ x=0.52 } text{or} boxed{ x=2.62 }$

(c) We would like to determine the solutions for the equation
$color{#4257b2}7sec x=7$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}sec x$ in the left side alone.

$$
7sec x=7
$$

$$
sec x=dfrac{7}{7}=1
$$

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

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

$$
cos x=1
$$

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

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

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

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

$$
x=0 text{or} x=2pi=6.28
$$

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

So the solutions of the equation are $boxed{ x=0 } text{or} boxed{ x=6.28 }$

(d) We would like to determine the solutions for the equation
$color{#4257b2}2csc x+17=15+csc x$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}csc x$ in the left side alone.

$$
2csc x+17=15+csc x
$$

$$
2csc x+17=15+csc x
$$

$$
2csc x+17-csc x-17=15+csc x-csc x-17
$$

$$
2csc x+cancel{17}-csc xcancel{-17}=15+cancel{csc x}cancel{-csc x}-17
$$

$$
csc x=-2
$$

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

$$
dfrac{1}{csc x}=dfrac{1}{-2}
$$

$$
sin x=-dfrac{1}{2}
$$

Now we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}sin^{-1}left(sin xright)=x$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{1}{2}right)$ not $color{#4257b2}sin^{-1}left(-dfrac{1}{2}right)$ 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 sine function for 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 our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ because we know that the sine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$$
x=dfrac{7pi}{6}=3.67 text{or} x=dfrac{11pi}{6}=5.76
$$

So the solutions of the equation are $boxed{ x=3.67 } text{or} boxed{ x=5.76 }$

(e) We would like to determine the solutions for the equation
$color{#4257b2}2sec x+1=6$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}sec x$ in the left side alone.

$$
2sec x+1=6
$$

$$
2sec x=6-1=5
$$

$$
sec x=dfrac{5}{2}
$$

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

$$
dfrac{1}{sec x}=dfrac{1}{dfrac{5}{2}}
$$

$$
cos x=dfrac{2}{5}
$$

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

$$
cos^{-1}left(cos xright)=cos^{-1}left(dfrac{2}{5}right)
$$

$$
x=cos^{-1}left(dfrac{2}{5}right)
$$

Note that $color{#4257b2}cos^{-1}left(cos xright)=x$. Now we can use the calculator to determine $color{#4257b2}cos^{-1}left(dfrac{2}{5}right)$ to find the acute angle.

$$
x=1.16
$$

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 of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos x=dfrac{2}{5}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $4$ because we know that the cosine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

$$
x=1.16 text{or} theta=2pi-1.16
$$

$$
x=1.16 text{or} x=5.12
$$

So the solutions of the equation are $boxed{ x=1.16 } text{or} boxed{ x=5.12 }$

(f) We would like to determine the solutions for the equation
$color{#4257b2}8+4cot x=10$, where $color{#4257b2}0 leq x leq 2pi$. First, we will make $color{#4257b2}cot x$ in the left side alone.

$$
8+4cot x=10
$$

$$
4cot x=10-8=2
$$

$$
cot x=dfrac{2}{4}=dfrac{1}{2}
$$

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

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

$$
tan x=2
$$

Now we can take $color{#4257b2}tan^{-1}$ for each side to find the value of the related acute angle.

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

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

Note that $color{#4257b2}tan^{-1}left(tan xright)=x$. Now we can use the calculator to determine $color{#4257b2}tan^{-1}left(2right)$ to find the acute angle.

$$
x=1.11
$$

Now we found the related acute angle for the equation $color{#4257b2}tan x=2$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}tan x=2$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ because we know that the tangent ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

$$
x=1.11 text{or} theta=pi+1.11
$$

$$
x=1.11 text{or} x=4.25
$$

So the solutions of the equation are $boxed{ x=1.11 } text{or} boxed{ x=4.25 }$

Large{$text{$text{$text{color{#c34632}(a) $x=0.79 text{radians} {color{Black}text{or}} x=3.93 text{radians}$
\
\
(b) $x=0.52 text{radians} {color{Black}text{or}} x=2.62 text{radians}$
\
\
Large{color{#c34632}(c) $x=0 text{radians} {color{Black}text{or}} x=6.28 text{radians}$
\
\
(d) $x=3.67 text{radians} {color{Black}text{or}} x=5.76 text{radians}$
\
\
Large{color{#c34632}(e) $x=1.16 text{radians} {color{Black}text{or}} x=5.12 text{radians}$
\
\
(f) $x=1.11 text{radians} {color{Black}text{or}} x=4.25 text{radians}$}$}$}$

(a) We would like to determine the solutions for the equation $color{#4257b2}sin 2x=dfrac{1}{sqrt{2}}$, where $color{#4257b2}0 leq x leq 2pi$. First, we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

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

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$, so $color{#4257b2}sin^{-1}left(sin 2xright)=2x$.

$$
2x=dfrac{pi}{4}
$$

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

Now we found the related acute angle for the equation $color{#4257b2}sin 2x=dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin 2x=dfrac{1}{sqrt{2}}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $2$ because we know that the sine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$$
2x=dfrac{pi}{4} text{or} 2x=dfrac{3pi}{4}
$$

$$
x=dfrac{pi}{8} text{or} x=dfrac{3pi}{8}
$$

But we know that the period of the function $color{#4257b2}sin 2x$ is $color{#4257b2}dfrac{2pi}{2}=pi$, so to find the remaining values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$ we can add the period $color{#4257b2}pi$ to the two solutions we found in the last step.

$$
x=dfrac{pi}{8}+pi text{or} x=dfrac{3pi}{8}+pi
$$

$$
x=dfrac{9pi}{8} text{or} x=dfrac{11pi}{8}
$$

Note that we added the period of the function only one time because if we added the period one more time the result will not be exist in the interval $color{#4257b2}0 leq x leq 2pi$.

So the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{8}, dfrac{3pi}{8}, dfrac{9pi}{8}, dfrac{11pi}{8}right}$

(b) We would like to determine the solutions for the equation $color{#4257b2}sin 4x=dfrac{1}{2}$, where $color{#4257b2}0 leq x leq 2pi$. First, we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

$$
sin 4x=dfrac{1}{2}
$$

$$
sin^{-1}left(sin 4xright)=sin^{-1}left(dfrac{1}{2}right)
$$

$$
4x=sin^{-1}left(dfrac{1}{2}right)
$$

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$, so $color{#4257b2}sin^{-1}left(sin 4xright)=4x$.

$$
4x=dfrac{pi}{6}
$$

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

Now we found the related acute angle for the equation $color{#4257b2}sin 4x=dfrac{1}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin 4x=dfrac{1}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $2$ because we know that the sine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

$$
4x=dfrac{pi}{6} text{or} 4x=pi-dfrac{pi}{6}
$$

$$
4x=dfrac{pi}{6} text{or} 4x=dfrac{5pi}{6}
$$

$$
x=dfrac{pi}{24} text{or} x=dfrac{5pi}{24}
$$

But we know that the period of the function $color{#4257b2}sin 4x$ is $color{#4257b2}dfrac{2pi}{4}=dfrac{pi}{2}$, so to find the remaining values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$ we can add the period $color{#4257b2}dfrac{pi}{2}$ to the two solutions we found in the last step.

$$
x=dfrac{pi}{24}+dfrac{pi}{2} text{or} x=dfrac{5pi}{24}+dfrac{pi}{2}
$$

$$
x=dfrac{13pi}{24} text{or} x=dfrac{17pi}{24}
$$

Now we can add the period $color{#4257b2}dfrac{pi}{2}$ to the two solutions we found in the last step.

$$
x=dfrac{13pi}{24}+dfrac{pi}{2} text{or} x=dfrac{17pi}{24}+dfrac{pi}{2}
$$

$$
x=dfrac{25pi}{24} text{or} x=dfrac{29pi}{24}
$$

Now we can add the period $color{#4257b2}dfrac{pi}{2}$ to the two solutions we found in the last step.

$$
x=dfrac{25pi}{24}+dfrac{pi}{2} text{or} x=dfrac{29pi}{24}+dfrac{pi}{2}
$$

$$
x=dfrac{37pi}{24} text{or} x=dfrac{41pi}{24}
$$

Note that we added the period of the function more than one time until we found all solutions in the interval $color{#4257b2}0 leq x leq 2pi$, so we added the period in this problem three times only because if we added the period one more time the result will not be exist in the interval $color{#4257b2}0 leq x leq 2pi$.

So the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{24}, dfrac{5pi}{24}, dfrac{13pi}{24}, dfrac{17pi}{24}, dfrac{25pi}{24}, dfrac{29pi}{24}, dfrac{37pi}{24}, dfrac{41pi}{24}right}$

(c) We would like to determine the solutions for the equation $color{#4257b2}sin 3x=-dfrac{sqrt{3}}{2}$, where $color{#4257b2}0 leq x leq 2pi$. First, we can take $color{#4257b2}sin^{-1}$ for each side to find the value of the related acute angle.

$$
sin 3x=-dfrac{sqrt{3}}{2}
$$

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

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

Note that $color{#4257b2}sin^{-1}left(sin thetaright)=theta$, so $color{#4257b2}sin^{-1}left(sin 3xright)=3x$. Now we will calculate $color{#4257b2}sin^{-1}left(dfrac{sqrt{3}}{2}right)$ not $color{#4257b2}sin^{-1}left(-dfrac{sqrt{3}}{2}right)$ to find the acute angle.

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

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

Now we found the related acute angle for the equation $color{#4257b2}sin 3x=-dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}sin 3x=-dfrac{sqrt{3}}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ because we know that the sine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

$$
3x=pi+dfrac{pi}{3}=dfrac{4pi}{3} text{or} 3x=2pi-dfrac{pi}{3}=dfrac{5pi}{3}
$$

$$
x=dfrac{4pi}{9} text{or} x=dfrac{5pi}{9}
$$

But we know that the period of the function $color{#4257b2}sin 3x$ is $color{#4257b2}dfrac{2pi}{3}$, so to find the remaining values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$ we can add the period $color{#4257b2}dfrac{2pi}{3}$ to the two solutions we found in the last step.

$$
x=dfrac{4pi}{9}+dfrac{2pi}{3} text{or} x=dfrac{5pi}{9}+dfrac{2pi}{3}
$$

$$
x=dfrac{10pi}{9} text{or} x=dfrac{11pi}{9}
$$

Now we can add the period $color{#4257b2}dfrac{2pi}{3}$ to the two solutions we found in the last step.

$$
x=dfrac{10pi}{9}+dfrac{2pi}{3} text{or} x=dfrac{11pi}{9}+dfrac{2pi}{3}
$$

$$
x=dfrac{16pi}{9} text{or} x=dfrac{17pi}{9}
$$

Note that we added the period of the function more than one time until we found all solutions in the interval $color{#4257b2}0 leq x leq 2pi$, so we added the period in this problem two times only because if we added the period one more time the result will not be exist in the interval $color{#4257b2}0 leq x leq 2pi$.

So the solutions of the equation are $color{#4257b2}x=left{dfrac{4pi}{9}, dfrac{5pi}{9}, dfrac{10pi}{9}, dfrac{11pi}{9}, dfrac{16pi}{9}, dfrac{17pi}{9}right}$

(d) We would like to determine the solutions for the equation $color{#4257b2}cos 4x=-dfrac{1}{sqrt{2}}$, where $color{#4257b2}0 leq x leq 2pi$. First, we can take $color{#4257b2}cos^{-1}$ for each side to find the value of the related acute angle.

$$
cos 4x=-dfrac{1}{sqrt{2}}
$$

$$
cos^{-1}left(cos 4xright)=cos^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

$$
4x=cos^{-1}left(-dfrac{1}{sqrt{2}}right)
$$

Note that $color{#4257b2}cos^{-1}left(cos thetaright)=theta$, so $color{#4257b2}cos^{-1}left(cos 4xright)=4x$. Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{sqrt{2}}right)$ not $color{#4257b2}cos^{-1}left(-dfrac{1}{sqrt{2}}right)$ to find the acute angle.

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

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

Now we found the related acute angle for the equation $color{#4257b2}cos 4x=-dfrac{1}{sqrt{2}}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos 4x=-dfrac{1}{sqrt{2}}$ which means that it is negative, so the solutions will be in quadrant $2$ and quadrant $3$ because we know that the cosine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$$
x=dfrac{3pi}{16} text{or} x=dfrac{5pi}{16}
$$

But we know that the period of the function $color{#4257b2}cos 4x$ is $color{#4257b2}dfrac{2pi}{4}=dfrac{pi}{2}$, so to find the remaining values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$ we can add the period $color{#4257b2}dfrac{pi}{2}$ to the two solutions we found in the last step.

$$
x=dfrac{3pi}{16}+dfrac{pi}{2} text{or} x=dfrac{5pi}{16}+dfrac{pi}{2}
$$

$$
x=dfrac{11pi}{16} text{or} x=dfrac{13pi}{16}
$$

Now we can add the period $color{#4257b2}dfrac{pi}{2}$ to the two solutions we found in the last step.

$$
x=dfrac{11pi}{16}+dfrac{pi}{2} text{or} x=dfrac{13pi}{16}+dfrac{pi}{2}
$$

$$
x=dfrac{19pi}{16} text{or} x=dfrac{21pi}{16}
$$

Now we can add the period $color{#4257b2}dfrac{pi}{2}$ to the two solutions we found in the last step.

$$
x=dfrac{19pi}{16}+dfrac{pi}{2} text{or} x=dfrac{21pi}{16}+dfrac{pi}{2}
$$

$$
x=dfrac{27pi}{16} text{or} x=dfrac{29pi}{16}
$$

Note that we added the period of the function more than one time until we found all solutions in the interval $color{#4257b2}0 leq x leq 2pi$, so we added the period in this problem three times only because if we added the period one more time the result will not be exist in the interval $color{#4257b2}0 leq x leq 2pi$.

So the solutions of the equation are $color{#4257b2}x=left{dfrac{3pi}{16}, dfrac{5pi}{16}, dfrac{11pi}{16}, dfrac{13pi}{16}, dfrac{19pi}{16}, dfrac{21pi}{16}, dfrac{27pi}{16}, dfrac{29pi}{16}right}$

(e) We would like to determine the solutions for the equation $color{#4257b2}cos 2x=-dfrac{1}{2}$, where $color{#4257b2}0 leq x leq 2pi$. First, we can take $color{#4257b2}cos^{-1}$ for each side to find the value of the related acute angle.

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

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

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

Note that $color{#4257b2}cos^{-1}left(cos thetaright)=theta$, so $color{#4257b2}cos^{-1}left(cos 2xright)=2x$. Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ not $color{#4257b2}cos^{-1}left(-dfrac{1}{2}right)$ to find the acute angle.

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

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

Now we found the related acute angle for the equation $color{#4257b2}cos 2x=-dfrac{1}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos 2x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $2$ and quadrant $3$ because we know that the cosine ratio is negative in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

$$
2x=pi-dfrac{pi}{3}=dfrac{2pi}{3} text{or} 2x=pi+dfrac{pi}{3}=dfrac{4pi}{3}
$$

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

But we know that the period of the function $color{#4257b2}cos 2x$ is $color{#4257b2}dfrac{2pi}{2}=pi$, so to find the remaining values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$ we can add the period $color{#4257b2}pi$ to the two solutions we found in the last step.

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

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

Note that we added the period of the function only one time because if we added the period one more time the result will not be exist in the interval $color{#4257b2}0 leq x leq 2pi$.

So the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{3}, dfrac{2pi}{3}, dfrac{4pi}{3}, dfrac{5pi}{3}right}$

(f) We would like to determine the solutions for the equation $color{#4257b2}cos dfrac{x}{2}=dfrac{sqrt{3}}{2}$, where $color{#4257b2}0 leq x leq 2pi$. First, we can take $color{#4257b2}cos^{-1}$ for each side to find the value of the related acute angle.

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

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

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

Note that $color{#4257b2}cos^{-1}left(cos thetaright)=theta$, so $color{#4257b2}cos^{-1}left(cos dfrac{x}{2}right)=dfrac{x}{2}$.

$$
dfrac{x}{2}=dfrac{pi}{6}
$$

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

Now we found the related acute angle for the equation $color{#4257b2}cos dfrac{x}{2}=dfrac{sqrt{3}}{2}$, so the next step is to know in which quadrants the solutions of our equation are exist and then use the acute angle to find these solutions. Since we know that $color{#4257b2}cos dfrac{x}{2}=dfrac{sqrt{3}}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $4$ because we know that the cosine ratio is positive in these two quadrants. Now we can use the acute angle to find the solutions in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$$
x=2cdot left(dfrac{pi}{6}right)=dfrac{pi}{3} text{or} x=2cdot left(dfrac{11pi}{6}right)=dfrac{11pi}{3}
$$

But we know that our interval is $color{#4257b2}0 leq x leq 2pi$, so the solution $color{#4257b2}dfrac{11pi}{3}$ is refused and the only solution is $color{#4257b2}x=dfrac{pi}{3}$.

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

Note that the period of the function is $color{#4257b2}dfrac{2pi}{dfrac{1}{2}}=4pi$, so we didn’t add this period to our solution because if we added the period the result will not be exist in the interval $color{#4257b2}0 leq x leq 2pi$.

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

$text{$text{color{#c34632}(a) $x=left{dfrac{pi}{8}, dfrac{3pi}{8}, dfrac{9pi}{8}, dfrac{11pi}{8}right}$ (b) $x=left{dfrac{pi}{24}, dfrac{5pi}{24}, dfrac{13pi}{24}, dfrac{17pi}{24}, dfrac{25pi}{24}, dfrac{29pi}{24}, dfrac{37pi}{24}, dfrac{41pi}{24}right}$
\
\
\
color{#c34632}(c) $x=left{dfrac{4pi}{9}, dfrac{5pi}{9}, dfrac{10pi}{9}, dfrac{11pi}{9}, dfrac{16pi}{9}, dfrac{17pi}{9}right}$ (d) $x=left{dfrac{3pi}{16}, dfrac{5pi}{16}, dfrac{11pi}{16}, dfrac{13pi}{16}, dfrac{19pi}{16}, dfrac{21pi}{16}, dfrac{27pi}{16}, dfrac{29pi}{16}right}$
\
\
\
Large{color{#c34632}(e) $x=left{dfrac{pi}{3}, dfrac{2pi}{3}, dfrac{4pi}{3}, dfrac{5pi}{3}right}$ (f) $x=left{dfrac{pi}{3}right}$}$}$

We would like to solve the equation $color{#4257b2}sin left(x+dfrac{pi}{4}right)=sqrt{2} cos x$ for $color{#4257b2}0 leq x leq 2pi$. First, we note the left side contains a sine function with a compound angle, so we can use the addition formula for the sine function to simplify where $color{#4257b2}sin (a+b)=sin acos b+cos asin b$.

$$
sin left(x+dfrac{pi}{4}right)=sqrt{2} cos x
$$

$$
sin xcos dfrac{pi}{4}+cos xsin dfrac{pi}{4}=sqrt{2} cos x
$$

$$
sin xcdot left(dfrac{sqrt{2}}{2}right)+cos xcdot left(dfrac{sqrt{2}}{2}right)=sqrt{2} cos x
$$

$$
dfrac{sqrt{2}}{2} sin x+dfrac{sqrt{2}}{2} cos x=sqrt{2} cos x
$$

Note that the angle $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the values of the sine and cosine function for it where $color{#4257b2}sin dfrac{pi}{4}=dfrac{sqrt{2}}{2}$ and $color{#4257b2}cos dfrac{pi}{4}=dfrac{sqrt{2}}{2}$. Now we can move $color{#4257b2}dfrac{sqrt{2}}{2} cos x$ to the other side by the opposite sign to simplify.

$$
dfrac{sqrt{2}}{2} sin x+dfrac{sqrt{2}}{2} cos x=sqrt{2} cos x
$$

$$
dfrac{sqrt{2}}{2} sin x=sqrt{2} cos x-dfrac{sqrt{2}}{2} cos x
$$

$$
dfrac{sqrt{2}}{2} sin x=dfrac{sqrt{2}}{2} cos x
$$

$$
cancel{dfrac{sqrt{2}}{2}} sin x=cancel{dfrac{sqrt{2}}{2}} cos x
$$

$$
sin x=cos x
$$

Now we can divide the two sides by $color{#4257b2}cos x$.

$$
dfrac{sin x}{cos x}=dfrac{sin x}{sin x}
$$

$$
tan x=1
$$

Note that we used the identity $color{#4257b2}tan x=dfrac{sin x}{cos x}$. Now we can take $color{#4257b2}tan^{-1}$ for each side to find the values of $color{#4257b2}x$.

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

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

But we know that $color{#4257b2}tan x$ is positive in quadrant $1$ and quadrant $3$, so we can find the values of $color{#4257b2}x$ where the reference angle is $color{#4257b2}dfrac{pi}{4}$.

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

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

$$text{color{#c34632}x=dfrac{pi}{4} {color{Black$text{or}} $x=dfrac{5pi}{4}$}}$

We would like to sketch the graph $color{#4257b2}y=sin 2theta$ for $color{#4257b2}0 leq theta leq 2pi$, so we can use the graphing calculator to sketch this graph.

Now we would like to indicate all solutions for the trigonometric equation $color{#4257b2}sin 2theta=-dfrac{1}{sqrt{2}}$. Since we have the graph of the function $color{#4257b2}y=sin 2theta$, so we can graph the line $color{#4257b2}y=-dfrac{1}{sqrt{2}}$ and find the points where the graph of the function $color{#4257b2}sin 2theta$ will intersect the line $color{#4257b2}y=-dfrac{1}{sqrt{2}}$ and then find the values of $color{#4257b2}theta$ at these points.

We note from the graph that the function $color{#4257b2}sin 2theta$ intersects the line $color{#4257b2}y=-dfrac{1}{sqrt{2}}$ at the points $color{#4257b2}A, B, C$ and $color{#4257b2}D$ where the values of $color{#4257b2}theta$ at these points are $color{#4257b2}dfrac{5pi}{8}, dfrac{7pi}{8}, dfrac{13pi}{8}$ and $color{#4257b2}dfrac{15pi}{8}$, so these are all solutions of the trigonometric function $color{#4257b2}sin 2theta=-dfrac{1}{sqrt{2}}$.

So all solutions of the equation $color{#4257b2}sin 2theta=-dfrac{1}{sqrt{2}}$ are $color{#4257b2}x=left{dfrac{5pi}{8}, dfrac{7pi}{8}, dfrac{13pi}{8}, dfrac{15pi}{8}right}$

$$
color{#c34632}x=left{dfrac{5pi}{8}, dfrac{7pi}{8}, dfrac{13pi}{8}, dfrac{15pi}{8}right}
$$

We would like to explain why the value of the function
$color{#4257b2}f(x)=25sin dfrac{pi}{50}left(x+20right)-55$ at $color{#4257b2}x=3$ is the same as the value of the function $color{#4257b2}x=7$. First, we will substitute the values of $color{#4257b2}x=3$ and $color{#4257b2}x=7$ in the function $color{#4257b2}f(x)$ to find the expressions of the function at each value and then simplify one expression of them to prove that it equals the second expression.

at $color{#4257b2}x=3$

$$
f(x)=25sin dfrac{pi}{50}left(x+20right)-55
$$

$$
f(3)=25sin dfrac{pi}{50}left(3+20right)-55
$$

$$
f(3)=25sin dfrac{pi}{50}left(23right)-55
$$

$$
boxed{ f(3)=25sin dfrac{23pi}{50}-55 }
$$

at $color{#4257b2}x=7$

$$
f(x)=25sin dfrac{pi}{50}left(x+20right)-55
$$

$$
f(7)=25sin dfrac{pi}{50}left(7+20right)-55
$$

$$
f(7)=25sin dfrac{pi}{50}left(27right)-55
$$

$$
boxed{ f(7)=25sin dfrac{27pi}{50}-55 }
$$

Now we found the expressions of the function at $color{#4257b2}x=3$ and $color{#4257b2}x=7$, so the next step is to simplify the first expression at $color{#4257b2}x=3$ to prove that it equals the second expression.

$$
f(3)=25sin dfrac{23pi}{50}-55
$$

But we know from the transformation that $color{#4257b2}sin left(pi-thetaright)=sin theta$, so we can use this identity to simplify our expression where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{23pi}{50}$ in our expression.

$$
f(3)=25sin left(pi-dfrac{23pi}{50}right)-55
$$

$$
f(3)=25sin left(dfrac{50pi}{50}-dfrac{23pi}{50}right)-55
$$

$$
f(3)=25sin dfrac{27pi}{50}-55
$$

Now we proved that the expression of the function at $color{#4257b2}x=3$ is the same expression at $color{#4257b2}x=7$, so we proved that the value of the function
$color{#4257b2}f(x)=25sin dfrac{pi}{50}left(x+20right)-55$ at $color{#4257b2}x=3$ is the same as the value of the function $color{#4257b2}x=7$.

$$
color{#c34632}f(3)=f(7)
$$

We would like to solve the trigonometric equation $color{#4257b2}2sin xcos x+sin x=0$. First, we note that all terms contain $color{#4257b2}sin x$, so we can take it as a common factor to simplify.

$$
2sin xcos x+sin x=0
$$

$$
sin xleft(2cos x+1right)=0
$$

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

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

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

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

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

We can take $color{#4257b2}sin^{-1}$ for each side to find the values of $color{#4257b2}x$.

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

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

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

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

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

Note that we found the values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x < 2pi$

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

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

We can take $color{#4257b2}cos^{-1}$ for each side to find the values of $color{#4257b2}x$ 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 $color{#4257b2}cos^{-1}left(cos xright)=x$. Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ not $color{#4257b2}cos^{-1}left(-dfrac{1}{2}right)$ 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 cosine function for it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we found the value of the related acute angle for $color{#4257b2}cos x=-dfrac{1}{2}$, so the next step is to find in which quadrants the solutions will be exist. We know that $color{#4257b2}cos x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $2$ and quadrant $3$ where the values of the cosine function is negative in these quadrants.

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

$$
x=dfrac{2pi}{3} text{or} x=dfrac{4pi}{3}
$$

Note that we found the values of $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x < 2pi$

So the solutions of the second case are $boxed{ x=dfrac{2pi}{3} } text{or} boxed{ x=dfrac{4pi}{3} }$

Now we found the values of $color{#4257b2}x$ in each case, so the solutions of the equation are $color{#4257b2}x=left{0, dfrac{2pi}{3}, pi, dfrac{4pi}{3}right}$

$$
color{#c34632}x=left{0, dfrac{2pi}{3}, pi, dfrac{4pi}{3}right}
$$

(a) We would like to solve the equation $color{#4257b2}sin 2x-2cos^{2}x=0$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that our equation contains $color{#4257b2}sin 2x$, so we can use the double angle formula for the sine function where $color{#4257b2}sin 2x=2sin xcos x$.

$$
sin 2x-2cos^{2}x=0
$$

$$
2sin xcos x-2cos^{2}x
$$

Now we note that all terms of our equation contain $color{#4257b2}2cos x$, so we can take it as a common factor to simplify.

$$
2cos xleft(sin x-cos xright)=0
$$

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

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

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

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 to find the values of $color{#4257b2}x$.

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

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

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

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

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 for 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 $color{#4257b2}sin x=cos x$

We can divide the two sides by $color{#4257b2}cos x$ to simplify.

$$
dfrac{sin x}{cos x}=dfrac{cos x}{cos x}
$$

$$
tan x=1
$$

Note that we used the identity $color{#4257b2}tan x=dfrac{sin x}{cos x}$. Now we can take $color{#4257b2}tan^{-1}$ for each side to find the values of $color{#4257b2}x$ to find the related acute angle.

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

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

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

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

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

Now we found the value of the related acute angle for $color{#4257b2}tan x=1$, so the next step is to find in which quadrants the solutions will be exist. We know that $color{#4257b2}tan x=1$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ where the values of the tangent function is positive in these quadrants.

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

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

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

Now we found the values of $color{#4257b2}x$ in each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{4}, dfrac{pi}{2}, dfrac{5pi}{4}, dfrac{3pi}{2}right}$

(b) We would like to solve the equation $color{#4257b2}3sin x+cos 2x=2$ for $color{#4257b2}0 leq x leq 2pi$. First, we note that our equation contains $color{#4257b2}cos 2x$, so we can use the double angle formula for the cosine function where $color{#4257b2}cos 2x=1-2sin^{2}x$.

$$
3sin x+cos 2x=2
$$

$$
3sin x+1-2sin^{2}x=2
$$

$$
-2sin^{2}x+3sin x+1=2
$$

$$
-2sin^{2}x+3sin x+1-2=0
$$

$$
-2sin^{2}x+3sin x-1=0
$$

Now we can multiply the equation by $-1$.

$$
2sin^{2}x-3sin x+1=0
$$

Now we note that we have a quadratic equation, so we can factor to simplify.

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

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

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

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

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

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

We can take $color{#4257b2}sin^{-1}$ for each side to find the values of $color{#4257b2}x$.

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

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

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

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

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

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

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

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

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

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

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

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

Now we found the value of the related acute angle for $color{#4257b2}sin x=dfrac{1}{2}$, so the next step is to find in which quadrants the solutions will be exist. We know that $color{#4257b2}sin x=dfrac{1}{2}$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $2$ where the values of the tangent function is positive in these quadrants.

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

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

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

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

$$
color{#c34632}(a) x=left{dfrac{pi}{4}, dfrac{pi}{2}, dfrac{5pi}{4}, dfrac{3pi}{2}right}
$$

$$
color{#c34632}(b) x=left{dfrac{pi}{6}, dfrac{pi}{2}, dfrac{5pi}{6}right}
$$