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

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

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

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

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