B2. Find all real numbers $x$ that satisfy the equation

$$
\log _{\sin x}\left(\frac{1}{2} \sin 2 x\right)=2
$$
B2. The base of the logarithm must be positive and different from 1, so $\sin x>0$ and $\sin x \neq 1$. Additionally, $\frac{1}{2} \sin 2 x>0$, since the logarithm is defined only for positive numbers. If this is true, we can transform the equation into an equivalent equation $\sin ^{2} x=\frac{1}{2} \sin 2 x=\sin x \cos x$ or $\sin x(\sin x-\cos x)=0$. Since $\sin x>0$, it follows that $\sin x-\cos x=0$. If $\cos x=0$, then $\sin x=0$ as well, which is not possible. Therefore, we can divide by $\cos x$ to get $\tan x=1$. From this, we get the solutions $x=\frac{\pi}{4}+2 k \pi$ and $x=\frac{5 \pi}{4}+2 k \pi$, where $k$ is an integer. However, $\sin \left(\frac{5 \pi}{4}+2 k \pi\right)=\sin \frac{5 \pi}{4}=-\frac{\sqrt{2}}{2}<0$ and $\sin x \neq 1$ ..... 1 point
Noting that $\frac{1}{2} \sin 2 x>0$ ..... 1 point
Transforming the equation to $\sin ^{2} x=\sin x \cos x$ ..... 1 point
Concluding that $\sin x-\cos x=0$ or $\tan x=1$ ..... 1 point
Writing only the correct solution to the equation ..... 2 points
(If the contestant does not exclude the incorrect solution, we will award only
1 point for writing the solutions.)