$6 \cdot 31$ Try to find the domain of the function: $y=\lg \left(\sin \frac{x}{2}-\frac{1}{2}\right)+\sqrt{\frac{1}{x-2}}$. (Shanghai Mathematical Competition, China, 1962)
[Solution] From the domain of $\lg \left(\sin \frac{x}{2}-\frac{1}{2}\right)$, we get $\sin \frac{x}{2}-\frac{1}{2}>0$, which means $\sin \frac{x}{2}>\frac{1}{2}$.

Thus, $\frac{\pi}{6}+2 k \pi < \frac{x}{2} < \frac{5 \pi}{6}+2 k \pi$, i.e., $\frac{\pi}{3}+4 k \pi < x < \frac{5 \pi}{3}+4 k \pi$.

Also, from $x-2>0$, we get $x>2$.
Therefore, the domain of $y$ is $\quad 2<x<\frac{5 \pi}{3}$.
and $\frac{\pi}{3}+4 k \pi<x<\frac{5 \pi}{3}+4 k \pi \cdot(k=1,2, \cdots)$