2. Solve the equation $\left(\frac{7}{4}-3 \cos 2 x\right) \cdot|1+2 \cos 2 x|=\sin x(\sin x+\sin 5 x)$.
Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in Z$.

Solution. Transform the right side of the equation:

$\sin x(\sin x+\sin 5 x)=\sin x \cdot 2 \sin 3 x \cos 2 x=\cos 2 x \cdot(2 \sin x \sin 3 x)=\cos 2 x \cdot(\cos 2 x-\cos 4 x)=$

$$
=\cos 2 x\left(-2 \cos ^{2} 2 x+\cos 2 x+1\right)=\cos 2 x(1-\cos 2 x)(2 \cos 2 x+1)
$$

Let $\cos 2 x=t$. Then the equation becomes $\left(\frac{7}{4}-3 t\right) \cdot|2 t+1|=t(2 t+1)(1-t)$. There are three cases.

a) $t=-\frac{1}{2}$ is a root of the equation.

b) $t>-\frac{1}{2}$. We get $\frac{7}{4}-3 t=t-t^{2}, t^{2}-4 t+\frac{7}{4}=0, t=\frac{1}{2}$ or $t=\frac{7}{2}$. $t=\frac{1}{2}$ is valid.

c) $t<-\frac{1}{2}$. We get $-\frac{7}{4}+3 t=t-t^{2}, t^{2}+2 t-\frac{7}{4}=0, t=\frac{-2 \pm \sqrt{11}}{2}$. The root $t_{1}=\frac{-2-\sqrt{11}}{2}$ is not valid, as $t_{1}<\therefore-1 ;$ the root $t_{2}=\frac{-2+\sqrt{11}}{2}$ is not valid, as it does not satisfy the condition $t<-\frac{1}{2}$.

In total: $t= \pm \frac{1}{2}, \cos 2 x= \pm \frac{1}{2}, 2 x= \pm \frac{\pi}{3}+k \pi, x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.