2. Solve the equation $\frac{\sin 5 x+\sin 7 x}{\sin 4 x-\sin 2 x}=-3 \cdot|\sin 2 x|$
Answer: $x=\pi-\arcsin \frac{\sqrt{57}-3}{8}+2 k \pi, x=\pi-\arcsin \frac{3-\sqrt{57}}{8}+2 k \pi, k \in \mathrm{Z}$.

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

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

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 $\cos x\left(3-4 \sin ^{2} x\right)=-3 \sin x \cos x$, from which either $\cos x=0$ (which is impossible, since the denominator in the left part of the original equation becomes zero), or $4 \sin ^{2} x-3 \sin x-3=0$. Therefore, $\sin x=\frac{3 \pm \sqrt{57}}{8}$. The equation $\sin x=\frac{3+\sqrt{57}}{8}$ has no solutions, as the right side is greater than one, and from the equation $\sin x=\frac{3-\sqrt{57}}{8}$, taking into account the restriction, we get $x=\pi-\arcsin \frac{3-\sqrt{57}}{8}+2 k \pi, k \in \mathrm{Z}$

b) $\sin x \cos x<0$ (i.e., the angle $x$ lies in the second or fourth quadrant). Then we get $\cos x\left(3-4 \sin ^{2} x\right)=3 \sin x \cos x$, from which $4 \sin ^{2} x+3 \sin x-3=0$. Therefore, $\sin x=\frac{-3 \pm \sqrt{57}}{8}$. The equation $\sin x=\frac{-3-\sqrt{57}}{8}$ has no solutions, as the right side is less than minus one, and from the equation $\sin x=\frac{\sqrt{57}-3}{8}$, taking into account the restriction, we get $x=\pi-\arcsin \frac{\sqrt{57}-3}{8}+2 k \pi, k \in \mathrm{Z}$