2. Solve the equation $\left(3 \cos 2 x+\frac{9}{4}\right) \cdot|1-2 \cos 2 x|=\sin x(\sin x-\sin 5 x)$.
Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.

Solution. Transform the right side of the equation:

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

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

Let $\cos 2 x=t$. Then the equation takes the form $\left(\frac{9}{4}+3 t\right) \cdot|1-2 t|=(1-2 t)\left(1-t^{2}\right)$. There are three cases.

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

b) $t<\frac{1}{2}$. We get $\frac{9}{4}+3 t=-(1-2 t)\left(1-t^{2}\right), \frac{9}{4}+3 t=-1+2 t-t^{2}+2 t^{3}, 2 t^{3}-t^{2}-5 t-\frac{13}{4}=0$. This cubic equation has no real roots that satisfy $t<\frac{1}{2}$.

c) $t>\frac{1}{2}$. We get $\frac{9}{4}+3 t=t^{2}-1, t^{2}-3 t-\frac{13}{4}=0, t=\frac{3 \pm \sqrt{22}}{2}$. The root $t_{1}=\frac{3+\sqrt{22}}{2}$ does not fit, as $t_{1}>1$; the root $t_{2}=\frac{3-\sqrt{22}}{2}$ does not fit, as it does not satisfy the condition $t>\frac{1}{2}$.

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