$8.447 \tan 2 x \cdot \tan^{2} 3 x \cdot \tan^{2} 5 x=\tan 2 x+\tan^{2} 3 x-\tan^{2} 5 x$.
## Solution.

$\operatorname{tg} 2 x \cdot\left(1-\operatorname{tg}^{2} 3 x \cdot \operatorname{tg}^{2} 5 x\right)=\operatorname{tg}^{2} 5 x-\operatorname{tg}^{2} 3 x \Leftrightarrow$

$\Leftrightarrow \frac{\sin 2 x}{\cos 2 x} \cdot\left(1-\frac{\sin 3 x \cdot \sin 5 x}{\cos 3 x \cdot \cos 5 x}\right)\left(1+\frac{\sin 3 x \cdot \sin 5 x}{\cos 3 x \cdot \cos 5 x}\right)=(\operatorname{tg} 5 x-\operatorname{tg} 3 x)(\operatorname{tg} 5 x+\operatorname{tg} 3 x) \Leftrightarrow$

$\Leftrightarrow \frac{\sin 2 x \cdot \cos (3 x+5 x) \cdot \cos (5 x-3 x)}{\cos 2 x \cdot \cos 3 x \cdot \cos 5 x}=\frac{\sin 2 x \cdot \sin 8 x}{\cos 3 x \cdot \cos 5 x}$.

$\left\{\begin{array}{l}\sin 2 x \cdot \cos 8 x=\sin 2 x \cdot \sin 8 x, \\ \cos 2 x \neq 0, \\ \cos 3 x \neq 0, \\ \cos 5 x \neq 0 .\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin 2 x \cdot(\cos 8 x-\sin 8 x)=0, \\ \cos 2 x \neq 0, \\ \cos 3 x \neq 0, \\ \cos 5 x \neq 0\end{array} \Leftrightarrow\right.\right.$

1) $\left\{\begin{array}{l}\sin 2 x=0, \\ \cos 3 x \neq 0, \\ \cos 5 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{\pi l}{2} \\ x \neq \frac{\pi}{6}+\frac{\pi n}{3} \\ x \neq \frac{\pi}{10}+\frac{\pi m}{5}\end{array}\right.\right.$

If $\frac{\pi l}{2}=\frac{\pi}{6}+\frac{\pi n}{3}$, then $l=1+2 t, t \in Z$; if $\frac{\dot{\pi} l}{2}=\frac{\pi}{10}+\frac{\pi m}{5}$, then $l=1+2 t, t \in Z \quad$ (see the method of solving equation (4)). Therefore, from the last system, it follows that $l$ is even, i.e., $l=2 k \Rightarrow x=\pi k$.

2) $\left\{\begin{array}{ll}\operatorname{tg} 8 x=1, \\ \cos 2 x \neq 0, \\ \cos 3 x \neq 0, \\ \cos 5 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{lr}x=\frac{\pi}{32}+\frac{\pi l}{8}, & \frac{\pi}{32}+\frac{\pi l}{8}=\frac{\pi}{4}+\frac{\pi t}{2} \\ x \neq \frac{\pi}{4}+\frac{\pi t}{2}, \\ x \neq \frac{\pi}{6}+\frac{\pi n}{3}, & \text { Since all three equations } \frac{\pi}{32}+\frac{\pi l}{8}=\frac{\pi}{6}+\frac{\pi n}{3} \\ x \neq \frac{\pi}{10}+\frac{\pi m}{5} . & \frac{\pi}{32}+\frac{\pi l}{8}=\frac{\pi}{10}+\frac{\pi m}{5}\end{array}\right.\right.$

do not have solutions (see the method of solving equation (4)), then $x=\frac{\pi}{32}+\frac{\pi l}{8}$.

Answer: $x_{1}=\pi k, x_{2}=\frac{\pi}{32}(4 k+1)$.

$8.448\left(5+\frac{3}{\sin ^{2} x}\right)\left(2-\sin ^{6} x\right)=7+\cos 2 y$.

Solution. Since $5+\frac{3}{\sin ^{2} x} \geq 8,2-\sin ^{6} x \geq 1$ for any $x$, the left side of the equation is not less than 8 for any $x$. Therefore, we have:

$\left\{\begin{array}{l}5+\frac{3}{\sin ^{2} x}=8, \\ 2-\sin ^{6} x=1, \\ 7+\cos 2 y=8\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin ^{2} x=1, \\ \cos 2 y=1\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{\pi}{2}+\pi k, \\ y=\pi m .\end{array}\right.\right.\right.$

Answer: $x=\frac{\pi}{2}(2 k+1), y=\pi m$.

## $8.449 \operatorname{tg}^{2} x+\operatorname{tg} x-3 \operatorname{ctg}^{2} x-3 \operatorname{ctg} x-2=0$

## Solution.

$\left\{\begin{array}{l}\left(\operatorname{tg}^{2} x+\operatorname{tg} x+1\right) \operatorname{tg}^{2} x-3\left(\operatorname{tg}^{2} x+\operatorname{tg} x+1\right)=0, \\ \operatorname{tg} x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\left(\operatorname{tg}^{2} x+\operatorname{tg} x+1\right)\left(\operatorname{tg}^{2} x-3\right)=0, \\ \operatorname{tg} x \neq 0\end{array} \Leftrightarrow\right.\right.$ $\Leftrightarrow \operatorname{tg} x= \pm \sqrt{3} \Leftrightarrow x= \pm \frac{\pi}{3}+\pi k$.

Answer: $\quad x=\frac{\pi}{3}(3 k \pm 1)$.