2. Solve the equation $\frac{\sin 8 x+\sin 4 x}{\cos 5 x+\cos x}=6|\sin 2 x|$.
Answer: $x=-\arccos \frac{\sqrt{13}-3}{4}+2 k \pi, x=-\arccos \frac{3-\sqrt{13}}{4}+2 k \pi, x=k \pi, k \in \mathrm{Z}$.

Solution. On the domain of definition, the given equation is equivalent to each of the following:

$\frac{2 \cos 2 x \sin 6 x}{2 \cos 3 x \cos 2 x}=6 \cdot|\sin 2 x| \Leftrightarrow \frac{2 \sin 3 x \cos 3 x}{\cos 3 x}=6 \cdot|\sin 2 x| \Leftrightarrow \sin 3 x=6 \cdot|\sin x \cos x| \Leftrightarrow \sin x\left(3-4 \sin ^{2} x\right)=6 \cdot|\sin x \cos x|$.

Consider two cases.

a) $\sin x \cos x \geq 0$ (i.e., the angle $x$ lies in the first or third quadrant). Then we get $\sin x\left(4 \cos ^{2} x-1\right)=6 \sin x \cos x$, from which either $\sin x=0$ and $x=k \pi, k \in \mathrm{Z}$, or $4 \cos ^{2} x-1=6 \cos x, 4 \cos ^{2} x-6 \cos x-1=0$. Therefore, $\cos x=\frac{3 \pm \sqrt{13}}{4}$. The equation $\cos x=\frac{3+\sqrt{13}}{4}$ has no solutions, as the right side is greater than one, and from the equation $\cos x=\frac{3-\sqrt{13}}{4}$, considering the restriction, we get $x=-\arccos \frac{3-\sqrt{13}}{4}+2 k \pi, k \in \mathrm{Z}$.

b) $\sin x \cos x<0$ (i.e., the angle $x$ lies in the second or fourth quadrant). Then we get $\sin x\left(4 \cos ^{2} x-1\right)=-6 \sin x \cos x$, from which $4 \cos ^{2} x+6 \cos x-1=0$. Therefore, $\cos x=\frac{-3 \pm \sqrt{13}}{4}$. The equation $\cos x=\frac{-3-\sqrt{13}}{4}$ has no solutions, as the right side is less than minus one, and from the equation $\cos x=\frac{-3+\sqrt{13}}{4}$, considering the restriction, we get $x=-\arccos \frac{\sqrt{13}-3}{4}+2 k \pi, k \in \mathrm{Z}$.