2. Find the domain of the function $y=\sqrt{\sin x-\frac{1}{2}}-\lg \left(\frac{1}{2}-\cos x\right)$,
$$
\begin{array}{l}
\sin x-\frac{1}{2} \geqslant 0, \\
\frac{1}{2}-\cos x>0
\end{array}
$$

The solution to the above inequalities is the same as the solution to:
$$
\begin{array}{l}
\sin x \geqslant \frac{1}{2}, \\
\cos x<\frac{1}{2}
\end{array}
$$

As shown in Figure 5, the solution to $\sin x \geqslant \frac{1}{2}$ is
$$
2 \mathrm{k} \pi+\frac{\pi}{6} \leqslant \mathrm{x} \leqslant 2 \mathrm{k} \pi+\frac{5 \pi}{6}, \quad \mathrm{k} \in \mathbb{Z},
$$
The solution to $\cos x<\frac{1}{2}$ is
$$
2 \mathrm{k} \pi+\frac{\pi}{3}<\mathrm{x}<2 \mathrm{k} \pi+\frac{5 \pi}{3}, \quad \mathrm{k} \in \mathbb{Z}.
$$

Therefore, the domain we are looking for is the intersection of these two solutions, which is
$$
2 \mathrm{k} \pi+\frac{\pi}{3}<\mathrm{x} \leqslant 2 \mathrm{k} \pi+\frac{5 \pi}{6}.
$$

From the above, we can see that using the unit circle and trigonometric line methods in problem-solving is highly intuitive and helps in discovering the relationship between known and unknown quantities. Therefore, we should strive to master this geometric problem-solving method. (Author's position: People's Education Press)