(a) We would like to state a trigonometric ratio that is equivalent to the trigonometric ratio $color{#4257b2}sin dfrac{3pi}{10}$. First, we know that $color{#4257b2}dfrac{3pi}{10}=pi-dfrac{7pi}{10}$, so we can rewrite our trigonometric ratio using this fact.

$$
sin dfrac{3pi}{10}=sin left(pi-dfrac{7pi}{10}right)
$$

Now we note that our trigonometric ratio is on the form $color{#4257b2}sin left(pi-thetaright)$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{7pi}{10}$ in our trigonometric ratio, so we can use the transformation where we know that $color{#4257b2}sin left(pi-thetaright)=sin theta$.

$$
sin dfrac{3pi}{10}=sin left(pi-dfrac{7pi}{10}right)=sin dfrac{7pi}{10}
$$

So the trigonometric ratio $color{#4257b2}sin dfrac{3pi}{10}$ can be written in the equivalent ratio $boxed{ sin dfrac{7pi}{10} }$

(b) We would like to state a trigonometric ratio that is equivalent to the trigonometric ratio $color{#4257b2}cos dfrac{6pi}{7}$. First, we know that $color{#4257b2}dfrac{6pi}{7}=pi-dfrac{pi}{7}$, so we can rewrite our trigonometric ratio using this fact.

$$
cos dfrac{6pi}{7}=cos left(pi-dfrac{pi}{7}right)
$$

Now we note that our trigonometric ratio is on the form $color{#4257b2}cos left(pi-thetaright)$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{pi}{7}$ in our trigonometric ratio, so we can use the transformation where we know that $color{#4257b2}cos left(pi-thetaright)=-cos theta$.

$$
cos dfrac{6pi}{7}=cos left(pi-dfrac{pi}{7}right)=-cos dfrac{pi}{7}
$$

So the trigonometric ratio $color{#4257b2}cos dfrac{6pi}{7}$ can be written in the equivalent ratio $boxed{ -cos dfrac{pi}{7} }$

(c) We would like to state a trigonometric ratio that is equivalent to the trigonometric ratio $color{#4257b2}-sin dfrac{13pi}{7}$. First, we know that $color{#4257b2}dfrac{13pi}{7}=2pi-dfrac{pi}{7}$, so we can rewrite our trigonometric ratio using this fact.

$$
-sin dfrac{13pi}{7}=-sin left(2pi-dfrac{pi}{7}right)
$$

Now we note that our trigonometric ratio is on the form $color{#4257b2}sin left(2pi-thetaright)$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{pi}{7}$ in our trigonometric ratio, so we can use the transformation where we know that $color{#4257b2}sin left(2pi-thetaright)=-sin theta$.

$$
-sin dfrac{13pi}{7}=-sin left(2pi-dfrac{pi}{7}right)=-left(-sin dfrac{pi}{7}right)=sin dfrac{pi}{7}
$$

So the trigonometric ratio $color{#4257b2}-sin dfrac{13pi}{7}$ can be written in the equivalent ratio $boxed{ sin dfrac{pi}{7} }$

(d) We would like to state a trigonometric ratio that is equivalent to the trigonometric ratio $color{#4257b2}-cos dfrac{8pi}{7}$. First, we know that $color{#4257b2}dfrac{8pi}{7}=pi+dfrac{pi}{7}$, so we can rewrite our trigonometric ratio using this fact.

$$
-cos dfrac{8pi}{7}=-cos left(pi+dfrac{pi}{7}right)
$$

Now we note that our trigonometric ratio is on the form $color{#4257b2}cos left(pi+thetaright)$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{pi}{7}$ in our trigonometric ratio, so we can use the transformation where we know that $color{#4257b2}cos left(pi+thetaright)=-cos theta$.

$$
-cos dfrac{8pi}{7}=-cos left(pi+dfrac{pi}{7}right)=-left(-cos dfrac{pi}{7}right)=cos dfrac{pi}{7}
$$

So the trigonometric ratio $color{#4257b2}-cos dfrac{8pi}{7}$ can be written in the equivalent ratio $boxed{ cos dfrac{pi}{7} }$

Large{$text{color{#c34632}(a) $sin dfrac{7pi}{10}$ (c) $sin dfrac{pi}{7}$
\
\
\
(b) $-cos dfrac{pi}{7}$ (d) $cos dfrac{pi}{7}$}$

We would like to write an equation that is equivalent to
$color{#4257b2}y=-5sin left(x-dfrac{pi}{2}right)-8$, using the cosine function. First, we can take $-1$ as a common factor from the angle of the sine function $color{#4257b2}x-dfrac{pi}{2}$.

$$
y=-5sin left(x-dfrac{pi}{2}right)-8
$$

$$
y=-5sin left[-left(dfrac{pi}{2}-xright)right]-8
$$

Now we note that our trigonometric ratio is on the form $color{#4257b2}sin left(-thetaright)$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{pi}{2}-x$ in our trigonometric ratio, so we can use the transformation where we know that $color{#4257b2}sin left(-thetaright)=-sin theta$.

$$
y=-5sin left[-left(dfrac{pi}{2}-xright)right]-8
$$

$$
y=5sin left(dfrac{pi}{2}-xright)-8
$$

Now we note that our trigonometric ratio is on the form $color{#4257b2}sin left(dfrac{pi}{2}-thetaright)$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{pi}{2}$ in our trigonometric ratio, so we can use the transformation where we know that $color{#4257b2}sin left(dfrac{pi}{2}-thetaright)=cos theta$.

$$
y=5sin left(dfrac{pi}{2}-xright)-8
$$

$$
y=5cos x-8
$$

So the equation $color{#4257b2}y=-5sin left(x-dfrac{pi}{2}right)-8$ can be written in the equivalent equation $boxed{ -y=5cos x-8 }$

Large{$text{color{#c34632}$y=5cos x-8$}$

(a) We would like to use a compound angle formula to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}sin left(x-dfrac{4pi}{3}right)$. First, we note that our expression is a sine function which consists of two subtraction angles, so we can use the subtraction formula of the sine function where $color{#4257b2}sin (a-b)=sin acos b-cos asin b$.

$$
begin{align*}
sin left(x-dfrac{4pi}{3}right)&=sin xcos dfrac{4pi}{3}-cos xsin dfrac{4pi}{3}
\ \
&=sin xcdot left(-dfrac{1}{2}right)-cos xcdot left(-dfrac{sqrt{3}}{2}right)
\ \
&=-dfrac{1}{2} sin x+dfrac{sqrt{3}}{2} cos x
\ \
&=dfrac{-sin x+sqrt{3} cos x}{2}
end{align*}
$$

Note that $color{#4257b2}dfrac{4pi}{3}$ is a special angle which we know the values of the sine and cosine functions of it where $color{#4257b2}sin dfrac{4pi}{3}=-dfrac{sqrt{3}}{2}$ and $color{#4257b2}cos dfrac{4pi}{3}=-dfrac{1}{2}$.

So the expression $color{#4257b2}sin left(x-dfrac{4pi}{3}right)$ can be written in the equivalent expression $boxed{ dfrac{-sin x+sqrt{3} cos x}{2} }$

(b) We would like to use a compound angle formula to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}cos left(x+dfrac{3pi}{4}right)$. First, we note that our expression is a cosine function which consists of two sum angles, so we can use the addition formula of the cosine function where $color{#4257b2}cos (a+b)=cos acos b-sin asin b$.

$$
begin{align*}
cos left(x+dfrac{3pi}{4}right)&=cos xcos dfrac{3pi}{4}-sin xsin dfrac{3pi}{4}
\ \
&=cos xcdot left(-dfrac{sqrt{2}}{2}right)-sin xcdot left(dfrac{sqrt{2}}{2}right)
\ \
&=-dfrac{sqrt{2}}{2} cos x-dfrac{sqrt{2}}{2} sin x
\ \
&=-dfrac{sqrt{2}}{2}left(cos x+sin xright)
end{align*}
$$

Note that $color{#4257b2}dfrac{3pi}{4}$ is a special angle which we know the values of the sine and cosine functions of it where $color{#4257b2}sin dfrac{3pi}{4}=dfrac{sqrt{2}}{2}$ and $color{#4257b2}cos dfrac{3pi}{4}=-dfrac{sqrt{2}}{2}$.

So the expression $color{#4257b2}cos left(x+dfrac{3pi}{4}right)$ can be written in the equivalent expression $boxed{ -dfrac{sqrt{2}}{2}left(cos x+sin xright) }$

(c) We would like to use a compound angle formula to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}tan left(x+dfrac{pi}{3}right)$. First, we note that our expression is a tangent function which consists of two sum angles, so we can use the addition formula of the tangent function where
$color{#4257b2}tan (a+b)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan left(x+dfrac{pi}{3}right)&=dfrac{tan x+tan dfrac{pi}{3}}{1-tan xtan dfrac{pi}{3}}
\ \
&=dfrac{tan x+sqrt{3}}{1-tan xcdot left(sqrt{3}right)}
\ \
&=dfrac{tan x+sqrt{3}}{1-sqrt{3} tan x}
end{align*}
$$

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

So the expression $color{#4257b2}tan left(x+dfrac{pi}{3}right)$ can be written in the equivalent expression $boxed{ dfrac{tan x+sqrt{3}}{1-sqrt{3} tan x} }$

(d) We would like to use a compound angle formula to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}cos left(x-dfrac{5pi}{4}right)$. First, we note that our expression is a cosine function which consists of two subtraction angles, so we can use the subtraction formula of the cosine function where $color{#4257b2}cos (a-b)=cos acos b+sin asin b$.

$$
begin{align*}
cos left(x-dfrac{3pi}{4}right)&=cos xcos dfrac{5pi}{4}+sin xsin dfrac{5pi}{4}
\ \
&=cos xcdot left(-dfrac{sqrt{2}}{2}right)+sin xcdot left(-dfrac{sqrt{2}}{2}right)
\ \
&=-dfrac{sqrt{2}}{2} cos x-dfrac{sqrt{2}}{2} sin x
\ \
&=-dfrac{sqrt{2}}{2}left(cos x+sin xright)
end{align*}
$$

Note that $color{#4257b2}dfrac{3pi}{4}$ is a special angle which we know the values of the sine and cosine functions of it where $color{#4257b2}sin dfrac{3pi}{4}=dfrac{sqrt{2}}{2}$ and $color{#4257b2}cos dfrac{3pi}{4}=-dfrac{sqrt{2}}{2}$.

So the expression $color{#4257b2}cos left(x-dfrac{5pi}{4}right)$ can be written in the equivalent expression $boxed{ -dfrac{sqrt{2}}{2}left(cos x+sin xright) }$

Large{$text{$text{color{#c34632}(a) $dfrac{-sin x+sqrt{3} cos x}{2}$ (c) $dfrac{tan x+sqrt{3}}{1-sqrt{3} tan x}$
\
\
\
Large{color{#c34632}(b) $-dfrac{sqrt{2}}{2}left(cos x+sin xright)$ (d) $-dfrac{sqrt{2}}{2}left(cos x+sin xright)$}$}$

(a) We would like to evaluate the expression $color{#4257b2}dfrac{tan dfrac{pi}{12}+tan dfrac{7pi}{4}}{1-tan dfrac{pi}{12}tan dfrac{7pi}{4}}$. First, we note that our expression is on the form of the addition formula of the tangent function where $color{#4257b2}tan (a+b)=dfrac{tan a+tan b}{1-tan atan b}$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{pi}{12}$ and $color{#4257b2}dfrac{7pi}{4}$ in our expression, so we can use this formula to evaluate our expression.

$$
begin{align*}
dfrac{tan dfrac{pi}{12}+tan dfrac{7pi}{4}}{1-tan dfrac{pi}{12}tan dfrac{7pi}{4}}&=tan left(dfrac{pi}{12}+dfrac{7pi}{4}right)
\ \
&=tan left(dfrac{pi}{12}+dfrac{21pi}{12}right)
\ \
&=tan dfrac{22pi}{12}
\ \
&=tan dfrac{11pi}{6}
\ \
&=-dfrac{1}{sqrt{3}}
end{align*}
$$

Note that $color{#4257b2}dfrac{11pi}{6}$ is a special angle which we know the value of the tangent function of it where $color{#4257b2}tan dfrac{11pi}{6}=-dfrac{1}{sqrt{3}}$.

So the expression $color{#4257b2}dfrac{tan dfrac{pi}{12}+tan dfrac{7pi}{4}}{1-tan dfrac{pi}{12}tan dfrac{7pi}{4}}$ equals $boxed{ -dfrac{1}{sqrt{3}} }$

(b) We would like to evaluate the expression $color{#4257b2}cos dfrac{pi}{9}cos dfrac{19pi}{18}-sin dfrac{pi}{9}sin dfrac{19pi}{18}$. First, we note that our expression is on the form of the addition formula of the cosine function where $color{#4257b2}cos (a+b)=cos acos b-sin asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{pi}{9}$ and $color{#4257b2}dfrac{19pi}{18}$ in our expression, so we can use this formula to evaluate our expression.

$$
begin{align*}
cos dfrac{pi}{9}cos dfrac{19pi}{18}-sin dfrac{pi}{9}sin dfrac{19pi}{18}&=cos left(dfrac{pi}{9}+dfrac{19pi}{18}right)
\ \
&=cos left(dfrac{2pi}{18}+dfrac{19pi}{18}right)
\ \
&=cos dfrac{21pi}{18}
\ \
&=cos dfrac{7pi}{6}
\ \
&=-dfrac{sqrt{3}}{2}
end{align*}
$$

Note that $color{#4257b2}dfrac{7pi}{6}$ is a special angle which we know the value of the cosine function of it where $color{#4257b2}cos dfrac{7pi}{6}=-dfrac{sqrt{3}}{2}$.

So the expression $color{#4257b2}cos dfrac{pi}{9}cos dfrac{19pi}{18}-sin dfrac{pi}{9}sin dfrac{19pi}{18}$ equals $boxed{ -dfrac{sqrt{3}}{2} }$

Large{$text{color{#c34632}(a) $-dfrac{1}{sqrt{3}}$ (b) $-dfrac{sqrt{3}}{2}$}$

(a) We would like to simplify the expression $color{#4257b2}2sin dfrac{pi}{12}cos dfrac{pi}{12}$. First, we note that our expression is on the form of the double angle formula of the sine function where $color{#4257b2}sin 2theta=2sin thetacos theta$ where $color{#4257b2}theta$ here is considered to be $color{#4257b2}dfrac{pi}{12}$ in our expression, so we can use this formula to simplify our expression.

$$
begin{align*}
2sin dfrac{pi}{12}cos dfrac{pi}{12}&=sin left(2cdot dfrac{pi}{12}right)
\ \
&=sin dfrac{pi}{6}
\ \
&=dfrac{1}{2}
end{align*}
$$

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

So the expression $color{#4257b2}2sin dfrac{pi}{12}cos dfrac{pi}{12}$ can be simplified to $boxed{ dfrac{1}{2} }$

(b) We would like to simplify the expression $color{#4257b2}cos^{2}dfrac{pi}{12}-sin^{2}dfrac{pi}{12}$. First, we note that our expression is on the form of the double angle formula of the cosine function where $color{#4257b2}cos 2theta=cos^{2}theta-sin^{2}theta$ where $color{#4257b2}theta$ here is considered to be $color{#4257b2}dfrac{pi}{12}$ in our expression, so we can use this formula to simplify our expression.

$$
begin{align*}
cos^{2}dfrac{pi}{12}-sin^{2}dfrac{pi}{12}&=cos left(2cdot dfrac{pi}{12}right)
\ \
&=cos dfrac{pi}{6}
\ \
&=dfrac{sqrt{3}}{2}
end{align*}
$$

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

So the expression $color{#4257b2}cos^{2}dfrac{pi}{12}-sin^{2}dfrac{pi}{12}$ can be simplified to $boxed{ dfrac{sqrt{3}}{2} }$

(c) We would like to simplify the expression $color{#4257b2}1-2sin^{2}dfrac{3pi}{8}$. First, we note that our expression is on the form of the double angle formula of the cosine function where $color{#4257b2}cos 2theta=1-2sin^{2}theta$ where $color{#4257b2}theta$ here is considered to be $color{#4257b2}dfrac{3pi}{8}$ in our expression, so we can use this formula to simplify our expression.

$$
begin{align*}
1-2sin^{2}dfrac{3pi}{8}&=cos left(2cdot dfrac{3pi}{8}right)
\ \
&=cos dfrac{3pi}{4}
\ \
&=cos left(pi-dfrac{pi}{4}right)
\ \
&=-cos dfrac{pi}{4}
\ \
&=-dfrac{1}{sqrt{2}}
end{align*}
$$

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

So the expression $color{#4257b2}1-2sin^{2}dfrac{3pi}{8}$ can be simplified to $boxed{ -dfrac{1}{sqrt{2}} }$

(d) We would like to simplify the expression $color{#4257b2}dfrac{2tan dfrac{pi}{6}}{1-tan^{2}dfrac{pi}{6}}$. First, we note that our expression is on the form of the double angle formula of the tangent function where $color{#4257b2}tan 2theta=dfrac{2tan theta}{1-tan^{2}theta}$ where $color{#4257b2}theta$ here is considered to be $color{#4257b2}dfrac{pi}{6}$ in our expression, so we can use this formula to simplify our expression.

$$
begin{align*}
dfrac{2tan dfrac{pi}{6}}{1-tan^{2}dfrac{pi}{6}}&=tan left(2cdot dfrac{pi}{6}right)
\ \
&=tan dfrac{pi}{3}
\ \
&=sqrt{3}
end{align*}
$$

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

So the expression $color{#4257b2}dfrac{2tan dfrac{pi}{6}}{1-tan^{2}dfrac{pi}{6}}$ can be simplified to $boxed{ sqrt{3} }$

Large{$text{$text{color{#c34632}(a) $dfrac{1}{2}$ (c) $-dfrac{1}{sqrt{2}}$
\
\
Large{color{#c34632}(b) $dfrac{sqrt{3}}{2}$ (d) $sqrt{3}$}$}$

(a) We would like to determine the values of $color{#4257b2}sin 2x, cos 2x$ and $color{#4257b2}tan 2x$ if we know that $color{#4257b2}sin x=dfrac{3}{5}$, and $color{#4257b2}x$ is acute. First, to determine the values of $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$ we need to know the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$ and then substitute in the double angle formulas of the sine and cosine functions. We note that $color{#4257b2}sin x=dfrac{3}{5}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find the value of $color{#4257b2}cos x$.

$$
sin^{2}x+cos^{2}x=1
$$

$$
left(dfrac{3}{5}right)^{2}+cos^{2}x=1
$$

$$
dfrac{9}{25}+cos^{2}x=1
$$

$$
cos^{2}x=1-dfrac{9}{25}=dfrac{16}{25}
$$

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

$$
cos x=pm sqrt{dfrac{16}{25}}=pm dfrac{4}{5}
$$

But we know that $color{#4257b2}x$ is acute angle which means that $color{#4257b2}x$ is in quadrant $1$, so the value of $color{#4257b2}cos x$ is positive and the negative solution is refused.

$$
boxed{ cos x=dfrac{4}{5} }
$$

Now we have the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$, so to determine $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$ we can use the double angle formulas of the sine and cosine function where $color{#4257b2}sin 2x=2sin xcos x$ and $color{#4257b2}cos 2x=cos^{2}x-sin^{2}x$.

$$
begin{align*}
sin 2x&=2sin xcos x
\ \
&=2cdot left(dfrac{3}{5}right)cdot left(dfrac{4}{5}right)
\ \
&=boxed{ dfrac{24}{25} }
end{align*}
$$

$$
begin{align*}
cos 2x&=cos^{2}x-sin^{2}x
\ \
&=left(dfrac{4}{5}right)^{2}-left(dfrac{3}{5}right)^{2}
\ \
&=dfrac{16}{25}-dfrac{9}{25}
\ \
&=boxed{ dfrac{7}{25} }
end{align*}
$$

Now we have the values of $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to determine $color{#4257b2}tan 2x$.

$$
begin{align*}
tan 2x&=dfrac{sin 2x}{cos 2x}
\ \
&=dfrac{dfrac{24}{25}}{dfrac{7}{25}}
\ \
&=dfrac{dfrac{24}{cancel{25}}}{dfrac{7}{cancel{25}}}
\ \
&=boxed{ dfrac{24}{7} }
end{align*}
$$

(b) We would like to determine the values of $color{#4257b2}sin 2x, cos 2x$ and $color{#4257b2}tan 2x$ if we know that $color{#4257b2}cot x=-dfrac{7}{24}$, and $color{#4257b2}x$ is obtuse. First, to determine the values of $color{#4257b2}tan 2x$ and $color{#4257b2}cos 2x$ we need to know the values of $color{#4257b2}tan x$ and $color{#4257b2}sin x$ and then substitute in the double angle formulas of the tangent and cosine functions. We note that $color{#4257b2}cot x=-dfrac{7}{24}$, so we can use the Pythagorean identity $color{#4257b2}1+cot^{2}theta=csc^{2}theta$ to find the value of $color{#4257b2}csc x$.

$$
1+cot^{2}x=csc^{2}x
$$

$$
1+left(-dfrac{7}{24}right)^{2}=csc^{2}x
$$

$$
csc^{2}x=1+dfrac{49}{576}=dfrac{625}{576}
$$

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

$$
csc x=pm sqrt{dfrac{625}{576}}=pm dfrac{25}{24}
$$

But we know that $color{#4257b2}x$ is obtuse angle which means that $color{#4257b2}x$ is in quadrant $2$, so the value of $color{#4257b2}csc x$ is positive and the negative solution is refused.

$$
csc x=dfrac{25}{24}
$$

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

$$
dfrac{1}{sin x}=dfrac{25}{24}
$$

$$
boxed{ sin x=dfrac{24}{25} }
$$

Now we can use the identity $color{#4257b2}tan theta=dfrac{1}{cot theta}$ to determine the value of $color{#4257b2}tan x$

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

$$
tan x=dfrac{1}{-dfrac{7}{24}}
$$

$$
boxed{ tan x=-dfrac{24}{7} }
$$

Now we have the values of $color{#4257b2}sin x$ and $color{#4257b2}tan x$, so to determine $color{#4257b2}cos 2x$ and $color{#4257b2}tan 2x$ we can use the double angle formulas of the cosine and tangent functions where $color{#4257b2}cos 2x=1-2sin^{2}x$ and $color{#4257b2}tan 2x=dfrac{2tan x}{1-tan^{2}x}$.

$$
begin{align*}
cos 2x&=1-2sin^{2}x
\ \
&=1-2cdot left(dfrac{24}{25}right)^{2}
\ \
&=1-2cdot left(dfrac{576}{625}right)=1-dfrac{1152}{625}
\ \
&=boxed{ -dfrac{527}{625} }
end{align*}
$$

$$
begin{align*}
tan 2x&=dfrac{2tan x}{1-tan^{2}x}
\ \
&=dfrac{2cdot left(-dfrac{24}{7}right)}{1-left(dfrac{24}{7}right)^{2}}
\ \
&=boxed{ dfrac{336}{527} }
end{align*}
$$

Now we have the values of $color{#4257b2}cos 2x$ and $color{#4257b2}tan 2x$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to determine $color{#4257b2}sin 2x$.

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

$$
sin 2x=tan 2xcos 2x
$$

$$
sin 2x=left(dfrac{336}{527}right)left(-dfrac{527}{625}right)
$$

$$
sin 2x=left(dfrac{336}{cancel{527}}right)left(-dfrac{cancel{527}}{625}right)
$$

$$
sin 2x=-dfrac{336}{625}
$$

(c) We would like to determine the values of $color{#4257b2}sin 2x, cos 2x$ and $color{#4257b2}tan 2x$ if we know that $color{#4257b2}cos x=dfrac{12}{13}$, and $color{#4257b2}dfrac{3pi}{2} leq x leq 2pi$. First, to determine the values of $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$ we need to know the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$ and then substitute in the double angle formulas of the sine and cosine functions. We note that $color{#4257b2}cos x=dfrac{12}{13}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find the value of $color{#4257b2}sin x$.

$$
sin^{2}x+cos^{2}x=1
$$

$$
sin^{2}x+left(dfrac{12}{13}right)^{2}=1
$$

$$
sin^{2}x+dfrac{144}{169}=1
$$

$$
sin^{2}x=1-dfrac{144}{169}=dfrac{25}{169}
$$

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

$$
sin x=pm sqrt{dfrac{25}{169}}=pm dfrac{5}{13}
$$

But we know that $color{#4257b2}dfrac{3pi}{2} leq x leq 2pi$ which means that $color{#4257b2}x$ is in quadrant $4$, so the value of $color{#4257b2}sin x$ is negative and the positive solution is refused.

$$
boxed{ sin x=-dfrac{5}{13} }
$$

Now we have the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$, so to determine $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$ we can use the double angle formulas of the sine and cosine function where $color{#4257b2}sin 2x=2sin xcos x$ and $color{#4257b2}cos 2x=cos^{2}x-sin^{2}x$.

$$
begin{align*}
sin 2x&=2sin xcos x
\ \
&=2cdot left(-dfrac{5}{13}right)cdot left(dfrac{12}{13}right)
\ \
&=boxed{ -dfrac{120}{169} }
end{align*}
$$

$$
begin{align*}
cos 2x&=cos^{2}x-sin^{2}x
\ \
&=left(dfrac{12}{13}right)^{2}-left(-dfrac{5}{13}right)^{2}
\ \
&=dfrac{144}{169}-dfrac{25}{169}
\ \
&=boxed{ dfrac{119}{169} }
end{align*}
$$

Now we have the values of $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to determine $color{#4257b2}tan 2x$.

$$
begin{align*}
tan 2x&=dfrac{sin 2x}{cos 2x}
\ \
&=dfrac{-dfrac{120}{169}}{dfrac{119}{169}}
\ \
&=dfrac{-dfrac{120}{cancel{169}}}{dfrac{119}{cancel{169}}}
\ \
&=boxed{ -dfrac{120}{119} }
end{align*}
$$

Large{$text{$text{$text{color{#c34632}(a) $sin 2x=dfrac{24}{25}, cos 2x=dfrac{7}{25}$ {color{Black}text{and}} $tan 2x=dfrac{24}{7}$
\
\
\
Large{color{#c34632}(b) $sin 2x=-dfrac{336}{625}, cos 2x=-dfrac{527}{625}$ {color{Black}text{and}} $tan 2x=dfrac{336}{527}$
\
\
\
Large{color{#c34632}(c) $sin 2x=-dfrac{120}{169}, cos 2x=dfrac{119}{169}$ {color{Black}text{and}} $tan 2x=-dfrac{120}{119}$}$}$}$

(a) We would like to determine if $color{#4257b2}tan 2x=dfrac{2sin xcos x}{1-2sin^{2}x}$ is a trigonometric equation or a trigonometric identity. First, we will simplify the right side and see it will equal the left side or not where if it equals the left side it will be a trigonometric identity and if not it will be a trigonometric equation.

We note that the right side is $color{#4257b2}dfrac{2sin xcos x}{1-2sin^{2}x}$ where the numerator is on the form of the double angle formula of the sine function where $color{#4257b2}sin 2theta=2sin thetacos theta$ and the denominator is on the form of the double angle formula of the cosine function where $color{#4257b2}cos 2theta=1-2sin^{2}theta$, so we can use these formulas to simplify the right side.

$$
begin{align*}
dfrac{2sin xcos x}{1-2sin^{2}x}&=dfrac{sin 2x}{cos 2x}
\ \
&=tan 2x
end{align*}
$$

Note that we used the identity $color{#4257b2}tan 2x=dfrac{sin 2x}{cos 2x}$. Now we note that the right side is simplified to $color{#4257b2}tan 2x$ which equals the left side, so $color{#4257b2}tan 2x=dfrac{2sin xcos x}{1-2sin^{2}x}$ $boxed{ text{is a trigonometric identity} }$

(b) We would like to determine if $color{#4257b2}sec^{2}x-tan^{2}x=cos x$ is a trigonometric equation or a trigonometric identity. First, we will simplify the left side and see it will equal the right side or not where if it equals the right side it will be a trigonometric identity and if not it will be a trigonometric equation.

We note that the left side is $color{#4257b2}sec^{2}x-tan^{2}x$, so we can use the Pythagorean identity $color{#4257b2}1+tan^{2}x=sec^{2}x$ to replace $color{#4257b2}sec^{2}x$ in the left side by $color{#4257b2}1+tan^{2}x$.

$$
begin{align*}
sec^{2}x-tan^{2}x&=1+tan^{2}x-tan^{2}x
\ \
&=1
end{align*}
$$

Now we note that the left side is simplified to $color{#4257b2}1$ which doesn’t equal the right side, so $color{#4257b2}sec^{2}x-tan^{2}x=cos x$ $boxed{ text{is a trigonometric equation} }$

(c) We would like to determine if $color{#4257b2}csc^{2}x-cot^{2}x=sin^{2}x+cos^{2}x$ is a trigonometric equation or a trigonometric identity. First, we will simplify the left side and then simplify the right side and see if they equal each others it will be a trigonometric identity and if not it will be a trigonometric equation.

We note that the left side is $color{#4257b2}csc^{2}x-cot^{2}x$, so we can use the Pythagorean identity $color{#4257b2}1+cot^{2}x=csc^{2}x$ to replace $color{#4257b2}csc^{2}x$ in the left side by $color{#4257b2}1+cot^{2}x$.

$$
begin{align*}
csc^{2}x-cot^{2}x&=1+cot^{2}x-cot^{2}x
\ \
&=1
end{align*}
$$

Now we note that the left side is simplified to $color{#4257b2}1$. Now we will simplify the right side.

We note that the right side is $color{#4257b2}sin^{2}x+cos^{2}x$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}x+cos^{2}x=1$.

$$
sin^{2}x+cos^{2}x=1
$$

Now we note that the right side is simplified to $color{#4257b2}1$ which equals the left side, so $color{#4257b2}csc^{2}x-cot^{2}x=sin^{2}x+cos^{2}x$ $boxed{ text{is a trigonometric identity} }$.

(d) We would like to determine if $color{#4257b2}tan^{2}x=1$ is a trigonometric equation or a trigonometric identity. First, we note that the left side is $color{#4257b2}tan^{2}x$ which doesn’t equal the right side except for some values of $color{#4257b2}x$, so $color{#4257b2}tan^{2}x=1$ $boxed{ text{is a trigonometric equation} }$

Large{$text{$text{color{#c34632}(a) Trigonometric identity (c) Trigonometric identity
\
\
Large{color{#c34632}(b) Trigonometric equation (d) Trigonometric equation}$}$

We would like to prove that $color{#4257b2}dfrac{1-sin^{2}x}{cot^{2}x}=1-cos^{2}x$ is a trigonometric identity. First, we will simplify the left side and see it will equal the left side or not.

We note that the left side is $color{#4257b2}dfrac{1-sin^{2}x}{cot^{2}x}$ but we know that the Pythagorean identity $color{#4257b2}sin^{2}x+cos^{2}x=1$, so $color{#4257b2}cos^{2}x=1-sin^{2}x$ and we can use this identity to replace $color{#4257b2}1-sin^{2}x$ from the numerator by $color{#4257b2}cos^{2}x$

$$
dfrac{1-sin^{2}x}{cot^{2}x}=dfrac{cos^{2}x}{cot^{2}x}
$$

Now we can use the identity $color{#4257b2}cot x=dfrac{cos x}{sin x}$ to replace $color{#4257b2}cot^{2}x$ from the denominator by $color{#4257b2}dfrac{cos^{2}x}{sin^{2}x}$.

$$
begin{align*}
dfrac{1-sin^{2}x}{cot^{2}x}&=dfrac{cos^{2}x}{cot^{2}x}
\ \
&=dfrac{cos^{2}x}{dfrac{cos^{2}x}{sin^{2}x}}
\ \
&=dfrac{cancel{cos^{2}x}}{dfrac{cancel{cos^{2}x}}{sin^{2}x}}
\ \
&=dfrac{1}{dfrac{1}{sin^{2}x}}
\ \
&=sin^{2}x
\ \
&=1-cos^{2}x
end{align*}
$$

Note that in the final step we used the Pythagorean identity one more time but this by replacing $color{#4257b2}sin^{2}x$ by $color{#4257b2}1-cos^{2}x$. Now we note that the left side can be simplified to $color{#4257b2}1-cos^{2}x$ which equals the right side, so
$color{#4257b2}dfrac{1-sin^{2}x}{cot^{2}x}=1-cos^{2}x$ is a trigonometric identity.

Large{$text{color{#c34632}$dfrac{1-sin^{2}x}{cot^{2}x}=1-cos^{2}x$}$

We would like to prove that $color{#4257b2}dfrac{2sec^{2}x-2tan^{2}x}{csc x}=sin 2xsec x$ is a trigonometric identity. First, we will simplify the left side and then simplify the right side and see if they will equal each others or not.

We note that the left side is $color{#4257b2}dfrac{2sec^{2}x-2tan^{2}x}{csc x}$ but we know that the Pythagorean identity $color{#4257b2}1+tan^{2}x=sec^{2}x$, so we can use this identity to replace $color{#4257b2}sec^{2}x$ from the numerator by $color{#4257b2}1+tan^{2}x$.

$$
begin{align*}
dfrac{2sec^{2}x-2tan^{2}x}{csc x}&=dfrac{2left(1+tan^{2}xright)-2tan^{2}x}{csc x}
\ \
&=dfrac{2+2tan^{2}x-2tan^{2}x}{csc x}
\ \
&=dfrac{2+cancel{2tan^{2}x}cancel{-2tan^{2}x}}{csc x}
\ \
&=dfrac{2}{csc x}
\ \
&=dfrac{2}{dfrac{1}{sin x}}
\ \
&=boxed{ 2sin x }
end{align*}
$$

Note that in the final step we used the identity $color{#4257b2}csc x=dfrac{1}{sin x}$. Now we note that the left side can be simplified to $color{#4257b2}2sin x$, so the next step is to simplify the right side and see if it will equal the left side or not.

We note that the right side is $color{#4257b2}sin 2xsec x$, so we can use the double angle formula for the sine function where $color{#4257b2}sin 2x=2sin xcos x$.

$$
begin{align*}
sin 2xsec x&=2sin xcos xsec x
\ \
&=2sin xcos xcdot left(dfrac{1}{cos x}right)
\ \
&=2sin xcancel{cos x}cdot left(dfrac{1}{cancel{cos x}}right)
\ \
&=boxed{ 2sin x }
end{align*}
$$

Note that in the second step we used the identity $color{#4257b2}sec x=dfrac{1}{cos x}$ to simplify. Now we note that the right side can be simplified also to $color{#4257b2}2sin x$, so the left side equals the right side and we proved that $color{#4257b2}dfrac{2sec^{2}x-2tan^{2}x}{csc x}=sin 2xsec x$ is a trigonometric identity.

Large{$text{color{#c34632}$dfrac{2sec^{2}x-2tan^{2}x}{csc x}=sin 2xsec x$}$

(a) We would like to solve the equation $color{#4257b2}dfrac{2}{sin x}+10=6$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we can subtract $color{#4257b2}10$ from each side to make $color{#4257b2}sin x$ in the left side alone.

$$
dfrac{2}{sin x}+10=6
$$

$$
dfrac{2}{sin x}+10-10=6-10
$$

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

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

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

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

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

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 determined 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 will be exist. We note that $color{#4257b2}sin x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

So the solutions of the equation are $boxed{ x=dfrac{7pi}{6} } text{or} boxed{ x=dfrac{11pi}{6} }$

(b) We would like to solve the equation $color{#4257b2}-dfrac{5cot x}{2}+dfrac{7}{3}=-dfrac{1}{6}$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we can subtract $color{#4257b2}dfrac{7}{3}$ from each side to make $color{#4257b2}cot x$ in the left side alone.

$$
-dfrac{5cot x}{2}+dfrac{7}{3}=-dfrac{1}{6}
$$

$$
-dfrac{5cot x}{2}+dfrac{7}{3}-dfrac{7}{3}=-dfrac{1}{6}-dfrac{7}{3}
$$

$$
-dfrac{5cot x}{2}=-dfrac{5}{2}
$$

Now we can multiply the two sides by $color{#4257b2}-dfrac{2}{5}$.

$$
left(-dfrac{2}{5}right)left(-dfrac{5cot x}{2}right)=left(-dfrac{2}{5}right)left(-dfrac{5}{2}right)
$$

$$
left(cancel{-dfrac{2}{5}}right)left(dfrac{cancel{-5}cot x}{cancel{2}}right)=left(cancel{-dfrac{2}{5}}right)left(cancel{-dfrac{5}{2}}right)
$$

$$
cot x=1
$$

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

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

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

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

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

Now we determined the related acute angle for the equation $color{#4257b2}cot x=1$, so the next step is to know in which quadrants the solutions will be exist. We note that $color{#4257b2}cot x=1$ which means that it is positive, so the solutions will be in quadrant $1$ and quadrant $3$ where the cotangent ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

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

(c) We would like to solve the equation $color{#4257b2}3+10sec x-1=-18$ in the interval $color{#4257b2}0 leq x leq 2pi$.

$$
3+10sec x-1=-18
$$

$$
2+10sec x=-18
$$

Now we can subtract $color{#4257b2}2$ from each side to make $color{#4257b2}sec x$ in the left side alone.

$$
2+10sec x-2=-18-2
$$

$$
10sec x=-20
$$

Now we can divide the two sides by $color{#4257b2}10$.

$$
dfrac{10sec x}{10}=dfrac{-20}{10}
$$

$$
sec x=-2
$$

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

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

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

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

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

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

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 determined 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 will be exist. We note 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 cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

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

$$
text{color{#c34632}(a) $x=dfrac{7pi}{6}$ {color{Black}text{or}} $x=dfrac{11pi}{6}$
\
\
\
Large{color{#c34632}(b) $x=dfrac{pi}{4}$ {color{Black}text{or}} $x=dfrac{5pi}{4}$}
\
\
\
Large{color{#c34632}(c) $x=dfrac{2pi}{3}$ {color{Black}text{or}} $x=dfrac{4pi}{3}$}}
$$

(a) We would like to solve the equation $color{#4257b2}y^{2}-4=0$. First, we can add $color{#4257b2}4$ to each side to make $color{#4257b2}y$ in the left side alone.

$$
y^{2}-4=0
$$

$$
y^{2}=4
$$

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

$$
y=pm sqrt{4}=pm 2
$$

So the solutions of the equation are $boxed{ y=2 } text{or} boxed{ y=-2 }$

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

$$
csc^{2}x-4=0
$$

$$
csc^{2}x=4
$$

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

$$
csc x=pm sqrt{4}=pm 2
$$

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

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

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

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

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

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

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

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

Now we determined 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 will be exist. We note 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 sine ratio is positive in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

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

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.

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

Now we determined 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 will be exist. We note that $color{#4257b2}sin x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

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

$$
text{color{#c34632}(a) $y=2$ {color{Black}text{or}} $y=-2$
\
\
\
Large{color{#c34632}(b) $x=left{dfrac{pi}{6}, dfrac{5pi}{6}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$}}
$$

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

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

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

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 determined 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 will be exist. We note that $color{#4257b2}sin x=-dfrac{1}{2}$ which means that it is negative, so the solutions will be in quadrant $3$ and quadrant $4$ where the sine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

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 of each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{2}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$

(b) We would like to solve the equation $color{#4257b2}tan^{2}xsin x-dfrac{sin x}{3}=0$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we note that all terms contain $color{#4257b2}sin x$, so we can take it as a common factor.

$$
tan^{2}xsin x-dfrac{sin x}{3}=0
$$

$$
sin xleft(tan^{2}x-dfrac{1}{3}right)=0
$$

Now we note that the term $color{#4257b2}tan^{2}x-dfrac{1}{3}$ is on the form $color{#4257b2}a^{2}-b^{2}$, so we can factor it where $color{#4257b2}a^{2}-b^{2}=left(a+bright)left(a-bright)$.

$$
sin xleft(tan x+dfrac{1}{sqrt{3}}right)left(tan x-dfrac{1}{sqrt{3}}right)=0
$$

Now we can use the zero-factor property.

$$
sin x=0, tan x+dfrac{1}{sqrt{3}}=0 text{or} tan x-dfrac{1}{sqrt{3}}=0
$$

$$
sin x=0, tan x=-dfrac{1}{sqrt{3}} text{or} tan x=dfrac{1}{sqrt{3}}
$$

Now we have three 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 $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$

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

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

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

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

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 determined 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 will be exist. We note that $color{#4257b2}tan x=-dfrac{1}{sqrt{3}}$ which means that it is negative, so the solutions will be 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 of the equation.

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

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

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

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 determined 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 will be exist. We note that $color{#4257b2}tan x=dfrac{1}{sqrt{3}}$ which means that it is positive, so the solutions will be 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 of the equation.

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

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

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

Now we found the solutions of each case, so the solutions of the equation are $color{#4257b2}x=left{0, dfrac{pi}{6}, dfrac{5pi}{6}, pi, dfrac{7pi}{6}, dfrac{11pi}{6}, 2piright}$

(c) We would like to solve the equation $color{#4257b2}cos^{2}x+left(dfrac{1-sqrt{2}}{2}right) cos x-dfrac{sqrt{2}}{4}=0$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we can multiply the two sides by $color{#4257b2}4$ to remove the denominator.

$$
cos^{2}x+left(dfrac{1-sqrt{2}}{2}right) cos x-dfrac{sqrt{2}}{4}=0
$$

$$
4cos^{2}x+2left(1-sqrt{2}right) cos x-sqrt{2}=0
$$

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

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

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

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

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

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

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

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

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

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

Now we determined the related acute angle for the equation $color{#4257b2}cos x=dfrac{sqrt{2}}{2}$, so the next step is to know in which quadrants the solutions will be exist. We note that $color{#4257b2}cos x=dfrac{sqrt{2}}{2}$ which means that it is positive, so the solutions will be 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 of the equation.

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

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

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

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

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

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

Now we will calculate $color{#4257b2}cos^{-1}left(dfrac{1}{2}right)$ to 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 of it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

Now we determined 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 will be exist. We note 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 cosine ratio is negative in these quadrants. Now we can use the related acute angle to find the solutions of the equation.

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

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

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 of each case, so the solutions of the equation are $color{#4257b2}x=left{dfrac{pi}{4}, dfrac{2pi}{3}, dfrac{4pi}{3}, dfrac{7pi}{4}right}$

(d) We would like to solve the equation $color{#4257b2}25tan^{2}x-70tan x=-49$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we can add $color{#4257b2}49$ to each side to make the right side equals zero.

$$
25tan^{2}x-70tan x=-49
$$

$$
25tan^{2}x-70tan x+49=0
$$

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

$$
left(5tan x-7right)left(5tan x-7right)=0
$$

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

$$
5tan x-7=0 text{or} 5tan x-7=0
$$

$$
tan x=dfrac{7}{5} text{or} tan x=dfrac{7}{5}
$$

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

$$
tan x=dfrac{7}{5}
$$

$$
tan^{-1}left(tan xright)=tan^{-1}left(dfrac{7}{5}right)
$$

$$
x=tan^{-1}left(dfrac{7}{5}right)
$$

$$
x=0.95 text{radians}
$$

Now we determined the related acute angle for the equation $color{#4257b2}tan x=dfrac{7}{5}$, so the next step is to know in which quadrants the solutions will be exist. We note that $color{#4257b2}tan x=dfrac{7}{5}$ which means that it is positive, so the solutions will be 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 of the equation.

$$
x=0.95 text{or} x=pi+0.95
$$

$$
x=0.95 text{radians} text{or} x=4.09 text{radians}
$$

So the solutions of the equation are $color{#4257b2}x=left{0.95 text{rdains}, 4.09 text{radians}right}$

$text{$text{color{#c34632}(a) $x=left{dfrac{pi}{2}, dfrac{7pi}{6}, dfrac{11pi}{6}right}$
\
\
\
Large{color{#c34632}(b) $x=left{0, dfrac{pi}{6}, dfrac{5pi}{6}, pi, dfrac{7pi}{6}, dfrac{11pi}{6}, 2piright}$}
\
\
\
Large{color{#c34632}(c) $x=left{dfrac{pi}{4}, dfrac{2pi}{3}, dfrac{4pi}{3}, dfrac{7pi}{4}right}$}
\
\
\
Large{color{#c34632}(d) $x=left{0.95 text{rdains}, 4.09 text{radians}right}$}$}$

We would like to solve the equation $color{#4257b2}dfrac{1}{1+tan^{2}x}=-cos x$ for $color{#4257b2}x$ in the interval $color{#4257b2}0 leq x leq 2pi$. First, we can use the Pythagorean identity $color{#4257b2}1+tan^{2}x=sec^{2}x$ to replace $color{#4257b2}1+tan^{2}x$ in the denominator in the left side by $color{#4257b2}sec^{2}x$ to simplify.

$$
dfrac{1}{1+tan^{2}x}=-cos x
$$

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

Now we can use the identity $color{#4257b2}sec x=dfrac{1}{cos x}$ to replace $color{#4257b2}sec^{2}x$ by $color{#4257b2}dfrac{1}{cos^{2}x}$.

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

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

Now we can add $color{#4257b2}cos x$ to each side to make the right side equals zero.

$$
cos^{2}x+cos x=0
$$

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

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

Now we can use the zero-factor property.

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

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

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

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

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

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

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

But we know that the range of the tangent function is $color{#4257b2}R-left{dfrac{pi}{2}+n piright}$ where $n$ is any integer number, so the solutions $color{#4257b2}x=dfrac{pi}{2} {color{Black}text{or}} x=dfrac{3pi}{2}$ are refused because the original equation contains $color{#4257b2}tan^{2}x$.

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 equation is $color{#4257b2}x=left{piright}$

Large{$text{color{#c34632}$x=left{piright}$}$