9.207. $\left\{\begin{array}{l}0,2^{\cos x} \leq 1, \\ \frac{x-1}{2-x}+\frac{1}{2}>0 .\end{array}\right.$
Solution.

Domain of definition: $x \neq 2$.

Rewrite the given system of inequalities as

$\left\{\begin{array}{l}0.2^{\cos x} \leq 0.2^{0}, \\ \frac{2 x-2+2-x}{2(2-x)}>0\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x \geq 0, \\ x(x-2)<0\end{array} \Leftrightarrow\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}-\frac{\pi}{2}+2 \pi n \leq x \leq \frac{\pi}{2}+2 \pi n, \\ 0<x<2 .\end{array} n \in Z\right.$. From this, we have $0<x \leq \frac{\pi}{2}$.

Answer: $x \in\left(0 ; \frac{\pi}{2}\right]$.