2. Solve the equation:

$$
\sin x=\sqrt{\sin ^{2} 3 x-\sin ^{2} 2 x}
$$
2. The given equation can be squared, with the condition $\sin x \geqslant 0$. Note that the expression under the root can be transformed as follows: $\sin ^{2} 3 x-\sin ^{2} 2 x=(\sin 3 x-\sin 2 x)(\sin 3 x+\sin 2 x)=2 \sin \frac{x}{2} \cos \frac{5 x}{2} \cdot 2 \sin \frac{5 x}{2} \cos \frac{x}{2}=$ $2 \sin \frac{x}{2} \cos \frac{x}{2} \cdot 2 \sin \frac{5 x}{2} \cos \frac{5 x}{2}=\sin x \sin 5 x$, and by squaring the given equation, we get $\sin ^{2} x=\sin x \sin 5 x$, which is equivalent to $\sin x(\sin 5 x-\sin x)=0$, i.e., $2 \sin x \sin 2 x \cos 3 x=0$. Therefore, we have $\sin x=0, \sin 2 x=0$ or $\cos 3 x=0$, i.e., $x=k \pi, x=\frac{k \pi}{2}$ or $x=\frac{\pi}{6}+\frac{k \pi}{3}$, for some $k \in \mathbb{Z}$. Taking into account the condition $\sin x \geqslant 0$, we obtain the following solutions:

$$
x \in\left\{k \pi, \frac{\pi}{2}+2 k \pi, \frac{\pi}{6}+2 k \pi, \frac{5 \pi}{6}+2 k \pi: k \in \mathbb{Z}\right\}
$$