1. Solve the equation $\frac{|\sin x|-\sin 3 x}{\cos x \cos 2 x}=2 \sqrt{3}$.
Answer. $x= \pm \frac{2 \pi}{3}+2 k \pi, x=-\frac{\pi}{6}+2 k \pi, k \in Z$.

Solution. There are two possible cases.

a) $\sin x \geq 0$. Then $\frac{\sin x-\sin 3 x}{\cos x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \frac{-2 \sin x \cos 2 x}{\cos x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \operatorname{tg} x=-\sqrt{3} \Leftrightarrow x=-\frac{\pi}{3}+\pi k, k \in Z . \quad$ Considering the condition, we get $x=\frac{2 \pi}{3}+2 k \pi, k \in Z$.

b) $\sin x<0$. Then $\frac{-\sin x-\sin 3 x}{\cos x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \frac{-2 \cos x \sin 2 x}{\cos x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \operatorname{tg} 2 x=-\sqrt{3} \Leftrightarrow x=-\frac{\pi}{6}+\frac{\pi k}{2}, k \in Z$. The sine is negative at $x=-\frac{2 \pi}{3}+2 k \pi, k \in Z$ and at $x=-\frac{\pi}{6}+2 k \pi, k \in Z$.