$8.482 \frac{3 \tan x-\tan^{3} x}{1-\tan^{2} x} \cdot(\cos 3 x+\cos x)=2 \sin 5 x$.
Solution. $\operatorname{tg} x \cdot \frac{3-\operatorname{tg}^{2} x}{1-\operatorname{tg}^{2} x} \cdot 2 \cos 2 x \cdot \cos x=2 \sin 5 x \Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{l}\sin x \cdot \cos 2 x \cdot \frac{3 \cos ^{2} x-\sin ^{2} x}{\cos ^{2} x-\sin ^{2} x}=\sin 5 x, \\ \cos x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x \cdot\left(3-4 \sin ^{2} x\right)=\sin 5 x, \\ \cos x \neq 0, \\ \cos 2 x \neq 0\end{array} \Leftrightarrow\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}\sin 3 x-\sin 5 x=0, \\ \cos x \neq 0, \\ \cos 2 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x \cdot \cos 4 x=0, \\ \cos x \neq 0, \\ \cos 2 x \neq 0 .\end{array} \Leftrightarrow\right.\right.$

1) $x=\pi k$.
2) $\left\{\begin{array}{l}\cos 4 x=0, \\ \cos x \neq 0, \\ \cos 2 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{\pi}{8}+\frac{\pi k}{4}, \\ x \neq \frac{\pi}{2}+\pi n, \\ x \neq \frac{\pi}{4}+\frac{\pi n}{2}\end{array} \Leftrightarrow x=\frac{\pi}{8}+\frac{\pi k}{4}\right.\right.$. (see the method of solving equation (4)).

Answer: $x_{1}=\pi k, x_{2}=\frac{\pi}{8}(2 k+1)$.