1. Solve the equation $\frac{|\sin x|+\sin 3 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}}$.
Answer. $x=\frac{\pi}{12}+2 k \pi, x=\frac{7 \pi}{12}+2 k \pi, x=-\frac{5 \pi}{6}+2 k \pi, k \in \mathrm{Z}$.

Solution. There are two possible cases.

a) $\sin x \geq 0$. Then $\frac{\sin x+\sin 3 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \frac{2 \cos x \sin 2 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \operatorname{tg} 2 x=\frac{1}{\sqrt{3}} \Leftrightarrow x=\frac{\pi}{12}+\frac{\pi k}{2}, k \in Z$. Considering the condition, we get $x=\frac{\pi}{12}+2 k \pi, k \in \mathrm{Z}$ and $x=\frac{7 \pi}{12}+2 k \pi, k \in \mathrm{Z}$.

b) $\sin x<0$. Then $\frac{-\sin x+\sin 3 x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \frac{2 \cos 2 x \sin x}{\cos x \cos 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \operatorname{tg} x=\frac{1}{\sqrt{3}} \Leftrightarrow x=\frac{\pi}{6}+k \pi, k \in Z$. The sine is negative when $x=-\frac{5 \pi}{6}+2 k \pi, k \in \mathrm{Z}$.