8.384. $\log _{\sin x} 4 \cdot \log _{\sin ^{2} x} 2=4$.
Solution.

Domain of definition: $0<\sin x<1$.

Switch to the base $\sin x$. We have

$$
\log _{\sin x} 2^{2} \cdot \frac{1}{2} \log _{\sin x} 2=4, \log _{\sin x}^{2} 2=4
$$

## Hence

1) $\log _{\sin x} 2=2, \sin ^{2} x=2, \varnothing$;
2) $\log _{\sin x} 2=-2, \frac{1}{\sin ^{2} x}=2, \sin ^{2} x=\frac{1}{2}, \sin x= \pm \frac{\sqrt{2}}{2}$.

Considering the domain of definition $\sin x=\frac{\sqrt{2}}{2}, x=(-1)^{k} \frac{\pi}{4}+\pi k, k \in Z$.

Answer: $x=(-1)^{k} \frac{\pi}{4}+\pi k, k \in Z$.