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

Solution. Transform the right side of the equation:

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

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

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

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

b) $t>\frac{1}{2}$. We get $\frac{7}{2} t+2=t^{2}+t, t^{2}-\frac{5}{2} t-2=0, t=\frac{5 \pm \sqrt{57}}{4}$. The root $t_{1}=\frac{5+\sqrt{57}}{4}$ does not fit, as $t_{1}>1$; the root $t_{2}=\frac{5-\sqrt{57}}{4}$ does not fit, as it does not satisfy the condition $t>\frac{1}{2}$.

c) $t<\frac{1}{2}$. We get $\frac{7}{2} t+2=-t^{2}-t, t^{2}+\frac{9}{2} t+2=0, t=4$ or $t=-\frac{1}{2}$. $t=-\frac{1}{2}$ fits.

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