7.244. $3 \log _{2}^{2} \sin x+\log _{2}(1-\cos 2 x)=2$.
Solution.

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

Since $1-\cos 2 x=2 \sin ^{2} x$, we have $3 \log _{2}^{2} \sin x+\log _{2} 2 \sin ^{2} x-2=0 \Leftrightarrow 3 \log _{2}^{2} \sin x+2 \log _{2} \sin x-1=0$.

Solving this equation as a quadratic in $\log _{2} \sin x$, we get $\log _{2} \sin x=\frac{1}{3}$ or $\log _{2} \sin x=-1$, from which $\sin x=\sqrt[3]{2}$ (no solutions), or $\sin x=\frac{1}{2}$. Then $x=(-1)^{n} \frac{\pi}{6}+\pi n, n \in Z$.

Answer: $(-1)^{n} \frac{\pi}{6}+\pi n$, where $n \in Z$.