$8.4743 \operatorname{tg} 2 x-4 \operatorname{tg} 3 x=\operatorname{tg}^{2} 3 x \cdot \operatorname{tg} 2 x$.
## Solution.

$$
\begin{aligned}
& 4(\operatorname{tg} 2 x-\operatorname{tg} 3 x)=\operatorname{tg}^{2} 3 x \cdot \operatorname{tg} 2 x+\operatorname{tg} 2 x \Leftrightarrow \frac{-4 \sin x}{\cos 2 x \cdot \cos 3 x}=\frac{\sin 2 x}{\cos 2 x}\left(1+\operatorname{tg}^{2} 3 x\right) \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \operatorname { cos } 2 x \neq 0 , } \\
{ \operatorname { cos } 3 x \neq 0 , } \\
{ 4 \operatorname { sin } x \cdot \operatorname { cos } 3 x = \operatorname { sin } 2 x }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\cos 2 x \neq 0, \\
\cos 3 x \neq 0, \\
-2 \sin x \cdot \cos 3 x=\cos x \cdot \sin x
\end{array}\right.\right. \\
& \left\{\begin{array} { l } 
{ \operatorname { cos } 2 x \neq 0 , } \\
{ \operatorname { cos } 3 x \neq 0 , } \\
{ [ \begin{array} { l } 
{ \operatorname { sin } x = 0 , } \\
{ 2 \operatorname { cos } 3 x + \operatorname { cos } x = 0 }
\end{array} }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\cos 2 x \neq 0, \\
\cos 3 x \neq 0, \\
{\left[\begin{array}{l}
x=\pi k, \\
8 \cos ^{3} x-5 \cos x=0 .
\end{array}\right.}
\end{array}\right.\right.
\end{aligned}
$$

1) $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \cos 3 x \neq 0, \\ x=\pi k\end{array} \Leftrightarrow x=\pi k\right.$.

$\int \cos 2 x \neq 0$

2) $\left\{\cos 3 x \neq 0, \quad \Leftrightarrow \cos ^{2} x=\frac{5}{8} \Leftrightarrow \operatorname{tg}^{2} x=\frac{3}{5} \Leftrightarrow x= \pm \operatorname{arctg} \sqrt{\frac{3}{5}}+\pi k\right.$. $[\cos x=0$, $8 \cos ^{2} x=5$

Answer: $\quad x_{1}=\pi k, x_{2}= \pm \operatorname{arctg} \sqrt{\frac{3}{5}}+\pi k$

## $8.475 \operatorname{ctg} 2 \pi t^{2}+\operatorname{ctg} 4 \pi t=0$.

## Solution.

$\frac{\sin \left(2 \pi t^{2}+4 \pi t\right)}{\sin 2 \pi t^{2} \cdot \sin 4 \pi t}=0 \Leftrightarrow\left\{\begin{array}{l}2 \pi t^{2}+4 \pi t=\pi k, k \in Z, \\ 2 \pi t^{2} \neq \pi l, l \in Z, \\ 4 \pi t \neq \pi n, n \in Z .\end{array}\right.$

If $2 t^{2}=l$, then $l+4 t=k \Rightarrow 4 t=k-l$ - integer. Therefore, the last system is equivalent to the system:

$\left\{\begin{array}{l}2 t^{2}+4 t-k=0, \\ 4 t \neq n, n \in Z\end{array} \Leftrightarrow\left\{\begin{array}{l}t=\frac{-2 \pm \sqrt{4+2 k}}{2} \\ 4 t \neq n, n \in Z .\end{array}, k=-1,0,1, \ldots\right.\right.$,

If $n=4 t$, then $n+4= \pm 2 \sqrt{4+2 k} \Rightarrow n^{2}+8 n=8 k \Rightarrow n^{2}: 8 \Rightarrow$ $n \vdots 4 \Rightarrow n=4 m \Rightarrow k=2 m(m+2)$. Or in a slightly different form:

$k=2\left(l^{2}-1\right)$ (if we put $m=l-1$ ).

Answer: $t=\frac{-2 \pm \sqrt{4+2 k}}{2}, k=-1,0,1,2, \ldots, k \neq 2\left(l^{2}-1\right), l \in Z$.

$8.476\left(3-\operatorname{tg}^{2} x\right)(\cos 3 x+\cos x)=\frac{4 \cos 3 x}{\operatorname{tg} 2 x}$.

Solution. $\left\{\begin{array}{l}\left(3-\frac{\sin ^{2} x}{\cos ^{2} x}\right)(\cos 3 x+\cos x)=\frac{4 \cos 3 x \cdot \cos 2 x}{\sin 2 x}, \\ \cos 2 x \neq 0\end{array} \Leftrightarrow\right.$

$$
\Leftrightarrow\left\{\begin{array} { l } 
{ \frac { ( 3 - 4 \operatorname { sin } ^ { 2 } x ) \operatorname { cos } 2 x \cdot \operatorname { cos } x } { \operatorname { cos } ^ { 2 } x } = \frac { 2 \operatorname { cos } 3 x \cdot \operatorname { cos } 2 x } { 2 \operatorname { sin } x \cdot \operatorname { cos } x } } \\
{ \operatorname { cos } 2 x \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \operatorname { cos } 2 x \neq 0 , } \\
{ \operatorname { sin } 2 x \neq 0 , } \\
{ \operatorname { sin } 3 x = \operatorname { cos } 3 x }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \neq \frac{\pi n}{4} \\
x=\frac{\pi}{12}+\frac{\pi k}{3}
\end{array}\right.\right.\right.
$$

If $\frac{\pi}{12}+\frac{\pi k}{3}=\frac{\pi n}{4}$, then $k=3 l+2, l \in Z$ (see the method of solving equation (4)).

Answer: $x=\frac{\pi}{12}(4 k+1), k \neq 3 l+2, l \in Z$.