8.330. $\operatorname{tg} 6 x \cos 2 x-\sin 2 x-2 \sin 4 x=0$.
Solution.

Domain of definition: $\cos 6 x \neq 0$.

Let's write the equation in the form

$\left(\frac{\sin 6 x \cos 2 x}{\cos 6 x}-\sin 2 x\right)-2 \sin 4 x=0$

$\frac{\sin 6 x \cos 2 x-\cos 6 x \sin 2 x}{\cos 6 x}-2 \sin 4 x=0 \Leftrightarrow \sin 4 x-2 \sin 4 x \cos 6 x=0$,

$\sin 4 x(1-2 \cos 6 x)=0$.

## From this

1) $\sin 4 x=0, 4 x=\pi k, x_{1}=\frac{\pi k}{4}, k \in Z$; considering the domain of definition, we get $=\frac{\pi m}{2}, m \in Z$
2) $1-2 \cos 6 x=0, \cos 6 x=\frac{1}{2}, 6 x= \pm \frac{\pi}{3}+2 \pi l, x_{2}= \pm \frac{\pi}{18}+\frac{\pi l}{3}, l \in Z$.

Answer: $x_{1}=\frac{\pi m}{2} ; x_{2}=\frac{\pi}{18}(6 l \pm 1)$, where $m$ and $l \in Z$.