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

Solution. Transform the right side of the equation:

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

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

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

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

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

c) $t<-\frac{1}{2}$. We get $\frac{7}{4}-2 t=t^{2}-1, t^{2}+2 t-\frac{11}{4}=0, t=\frac{-2 \pm \sqrt{15}}{2}$. The root $t_{1}=\frac{-2-\sqrt{15}}{2}$ is not valid, as $t_{1}<-1$; the root $t_{2}=\frac{-2+\sqrt{15}}{2}$ is not valid, as it does not satisfy the condition $t<-\frac{1}{2}$.

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