9.134. $0.5^{\sqrt{3}}<0.5^{\frac{\sin 2 x}{1-\cos 2 x}}<0.5$.
## Solution.

The given inequality is equivalent to the system of two inequalities

$\left\{\begin{array}{l}\frac{\sin 2 x}{1-\cos 2 x}>1, \\ \frac{\sin 2 x}{1-\cos 2 x}>0, \\ \frac{\sin 2 x}{1-\cos 2 x}-\sqrt{3}>0, \\ \frac{\sin 2 x-\sqrt{3}+\sqrt{3} \cos 2 x}{1-\cos 2 x}>0, \\ \frac{2 \sin x \cos x-\sqrt{3} \cos ^{2} x-\sqrt{3} \sin ^{2} x+\sqrt{3} \cos ^{2} x-\sqrt{3} \sin ^{2} x}{1-\cos ^{2} x+\sin ^{2} x}>0, \\ \frac{2 \sin x(\cos x-\sqrt{3} \sin x)}{2 \sin ^{2} x}>0, \\ \operatorname{ctg} x-\sqrt{3}>1, \\ \operatorname{ctg} x<\sqrt{3}, \\ \sin x \neq 0\end{array} \Leftrightarrow\right.\right.\right.$ $\Leftrightarrow 1<\operatorname{ctg} x<\sqrt{3} \Leftrightarrow \frac{\pi}{6}+\pi k<x<\frac{\pi}{4}+\pi k, k \in Z$.

Answer: $x \in\left(\frac{\pi}{6}+\pi k ; \frac{\pi}{4}+\pi k\right), k \in Z$.