1. Solve the equation $\frac{|\cos x|-\cos 3 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}}$.
Answer. $x=\frac{\pi}{6}+2 k \pi, x=\frac{5 \pi}{6}+2 k \pi, x=\frac{4 \pi}{3}+2 k \pi, k \in Z$.

Solution. There are two possible cases.

a) $\cos x \geq 0$. Then $\frac{\cos x-\cos 3 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \frac{2 \sin x \sin 2 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \operatorname{tg} x=\frac{1}{\sqrt{3}} \Leftrightarrow x=\frac{\pi}{6}+\pi k, k \in Z$. Considering the condition, we get $x=\frac{\pi}{6}+2 k \pi, k \in Z$.

b) $\cos x<0$. Then $\frac{-\cos x-\cos 3 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \frac{-2 \cos x \cos 2 x}{\cos x \sin 2 x}=\frac{2}{\sqrt{3}} \Leftrightarrow \operatorname{ctg} 2 x=-\frac{1}{\sqrt{3}} \Leftrightarrow x=\frac{\pi}{3}+\frac{\pi k}{2}, k \in Z$. The cosine is negative for $x=\frac{4 \pi}{3}+2 k \pi, k \in Z$ and for $x=\frac{5 \pi}{6}+2 k \pi, k \in Z$.