2. Solve the equation $\frac{\cos 5 x-\cos 7 x}{\sin 4 x+\sin 2 x}=2|\sin 2 x|$.
Answer: $x=-\arccos \frac{\sqrt{13}-1}{4}+2 k \pi, x=-\arccos \frac{1-\sqrt{13}}{4}+2 k \pi, k \in Z$.

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

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

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 $\sin x\left(4 \cos ^{2} x-3\right)=2 \sin x \cos x$, from which either $\sin x=0$ (which is impossible, as the denominator in the left part of the original equation becomes zero), or $4 \cos ^{2} x-2 \cos x-3=0$. Therefore, $\cos x=\frac{1 \pm \sqrt{13}}{4}$. The equation $\cos x=\frac{1+\sqrt{13}}{4}$ has no solutions, as the right side is greater than one, and from the equation $\cos x=\frac{1-\sqrt{13}}{4}$, taking into account the restriction, we get $x=-\arccos \frac{1-\sqrt{13}}{4}+2 k \pi, k \in Z$.

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