1. Solve the equation $\frac{3 \sin 3 x}{\sin x}-\frac{2 \cos 3 x}{\cos x}=7|\cos x|$.
Answer: $x= \pm \arccos \frac{3}{4}+k \pi, k \in Z$.

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

$$
\frac{3\left(3 \sin x-4 \sin ^{3} x\right)}{\sin x}-\frac{2\left(4 \cos ^{3} x-3 \cos x\right)}{\cos x}=7|\cos x| \Leftrightarrow 9-12 \sin ^{2} x-8 \cos ^{2} x+6=7|\cos x| \Leftrightarrow 4 \cos ^{2} x-7|\cos x|+3=0
$$

Solving the last equation as a quadratic equation in terms of $|\cos x|$, we get that $|\cos x|=1$ or $|\cos x|=\frac{3}{4}$. The first case does not satisfy the domain of definition (since if $|\cos x|=1$, then $\sin x=0$). From the second equation, we find that $x= \pm \arccos \frac{3}{4}+k \pi, k \in Z$.