$8.456 \tan^{2} x \cdot \tan^{2} 3 x \cdot \tan 4 x=\tan^{2} x-\tan^{2} 3 x+\tan 4 x$.
## Solution.

$\operatorname{tg} 4 x \cdot\left(1-\operatorname{tg}^{2} x \cdot \operatorname{tg}^{2} 3 x\right)=(\operatorname{tg} 3 x-\operatorname{tg} x)(\operatorname{tg} 3 x+\operatorname{tg} x) \Leftrightarrow$ $\Leftrightarrow \frac{\sin 4 x}{\cos 4 x} \cdot\left(1-\frac{\sin x \cdot \sin 3 x}{\cos x \cdot \cos 3 x}\right)\left(1+\frac{\sin x \cdot \sin 3 x}{\cos x \cdot \cos 3 x}\right)=\frac{\sin 2 x \cdot \sin 4 x}{\cos ^{2} x \cdot \cos ^{2} 3 x} \Leftrightarrow$ $\Leftrightarrow \frac{\sin 4 x \cdot \cos 4 x \cdot \cos 2 x}{\cos 4 x \cdot \cos x \cdot \cos 3 x}=\frac{\sin 2 x \cdot \sin 4 x}{\cos x \cdot \cos 3 x} \Leftrightarrow$

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

1) $\left\{\begin{array}{l}\sin 4 x=0, \\ \cos x \neq 0, \\ \cos 3 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{\pi l}{4}, \\ x \neq \frac{\pi}{2}+\pi n, \\ x \neq \frac{\pi}{6}+\frac{\pi m}{3} .\end{array}\right.\right.$

If $\frac{\pi l}{4}=\frac{\pi}{2}+\pi n$, then $l=2+4 n, n \in Z$;

if $\frac{\pi l}{4}=\frac{\pi}{6}+\frac{\pi m}{3}$, then $l=2+4 t, t \in Z$ (see the method of solving equation (4)).

Therefore, $x=\frac{\pi l}{4}, l \neq 2+4 k, k \in Z$ or $x_{1}=\pi k, x_{2}=\frac{\pi}{4}+\frac{\pi k}{2}$.

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

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