$8.444 \frac{\tan x}{2-\frac{1}{\cos ^{2} x}} \cdot(\sin 3 x-\sin x)=\frac{2}{\cot^{2} x-3}$.
Solution. The original equation is equivalent to the following system:

$$
\left\{\begin{array}{l}
\frac{\sin x \cdot \cos x}{2 \cos ^{2} x-1} \cdot 2 \sin x \cdot \cos 2 x=\frac{2 \sin ^{2} x}{\cos ^{2} x-3 \sin ^{2} x}, \\
\cos x \neq 0, \\
\sin x \neq 0
\end{array} \Leftrightarrow\right.
$$

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

Answer: $x= \pm \frac{2 \pi}{3}+2 \pi k$.

## $8.445 \operatorname{tg}(\pi \cos x)=\operatorname{ctg}(\pi \sin x)$

Solution. $\operatorname{tg}(\pi \cos x)=\operatorname{tg}\left(\frac{\pi}{2}-\pi \sin x\right) \Leftrightarrow\left\{\begin{array}{l}\pi \cos x \neq \frac{\pi}{2}+\pi t, t \in Z, \\ \pi \sin x \neq \pi l, l \in Z, \\ \pi \cos x-\frac{\pi}{2}+\pi \sin x=\pi n, n \in Z\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}\cos x \neq \frac{1}{2}+t \\ \sin x \neq l \\ \cos x+\sin x=\frac{1}{2}+n\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x \neq 0, \pm 1 \\ \cos x+\sin x=\frac{1}{2}+n .\end{array}\right.\right.$

Since from $\sin x=0$ it follows that $\cos x= \pm 1$; and from $\sin x= \pm 1$ it follows that $\cos x=0$, the last system is equivalent to the following equation:

$\cos x+\sin x=\frac{1}{2}+n \Leftrightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{1+2 n}{2 \sqrt{2}}, n=0,-1 \Leftrightarrow$

$\Leftrightarrow x=\frac{\pi}{4} \pm \arccos \left( \pm \frac{\sqrt{2}}{4}\right)+2 \pi k=\frac{\pi}{4} \pm \arccos \frac{\sqrt{2}}{4}+\pi k$.

Answer: $\quad x=\frac{\pi}{4} \pm \arccos \frac{\sqrt{2}}{4}+\pi k$.