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

Solution. There are two possible cases.

a) $\cos x \geq 0$. Then $\frac{\cos x+\cos 3 x}{\sin x \cos 2 x}=-2 \sqrt{3} \Leftrightarrow \frac{2 \cos x \cos 2 x}{\sin x \cos 2 x}=-2 \sqrt{3} \Leftrightarrow \operatorname{ctg} x=-\sqrt{3} \Leftrightarrow x=-\frac{\pi}{6}+\pi k, k \in Z$. Considering the condition, we get $x=-\frac{\pi}{6}+2 k \pi, k \in Z$.

b) $\quad \cos x<0$. Then $\frac{-\cos x+\cos 3 x}{\sin x \cos 2 x}=2 \sqrt{3} \Leftrightarrow \frac{-2 \sin x \sin 2 x}{\sin x \cos 2 x}=-2 \sqrt{3} \Leftrightarrow \operatorname{tg} 2 x=\sqrt{3} \Leftrightarrow x=\frac{\pi}{6}+\frac{\pi k}{2}, k \in Z$. The cosine is negative for $x=\frac{2 \pi}{3}+2 k \pi, k \in \mathrm{Z}$ and for $x=\frac{7 \pi}{6}+2 k \pi, k \in \mathrm{Z}$.