$5.31 \operatorname{tg} 6 x \cos 2 x-\sin 2 x-2 \sin 4 x=0$.
5.31 Here $\cos 6 x \neq 0$. Therefore,

$\sin 6 x \cos 2 x - \sin 2 x \cos 6 x = 2 \sin 4 x \cos 6 x$

$\sin 4 x = 2 \sin 4 x \cos 6 x$
from which we get:

1) $\sin 4 x = 0 ; x = \frac{\pi k}{4}$, but for $k = 2 l + 1$ we have $\cos 6 x = 0$, so $x_{1} = \frac{\pi k}{2}$
2) $\cos 6 x = \frac{1}{2} ; x_{2} = \frac{\pi}{18}(6 k \pm 1)$.

Answer: $x_{1} = \frac{\pi k}{2}, x_{2} = \frac{\pi}{18}(6 k \pm 1), k \in Z$.