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

Solution. The given equation, under the condition $\cos x \sin x \neq 0$, is equivalent to each of the following:

$$
\frac{\left(3 \sin x-4 \sin ^{3} x\right)}{\sin x}-\frac{2\left(4 \cos ^{3} x-3 \cos x\right)}{\cos x}=5|\sin x| \Leftrightarrow 3-4 \sin ^{2} x-8 \cos ^{2} x+6=5|\sin x| \Leftrightarrow 4 \sin ^{2} x-5|\sin x|+1=0
$$

Solving the last equation as a quadratic in terms of $|\sin x|$, we get that $|\sin x|=1$ or $|\sin x|=\frac{1}{4}$. The first case does not satisfy the domain of definition (since if $|\sin x|=1$, then $\cos x=0$). From the second equation, we find that $x= \pm \arcsin \frac{1}{4}+k \pi, k \in \mathrm{Z}$.