Example 5. Express $x$ in terms of the inverse sine function for the following equation: $\sin x=\frac{1}{3}, x \in(5 \pi, 8 \pi)$.
$$
\begin{array}{l}
\text { Brief solution: Divide the interval }(5 \pi, 8 \pi) \text { into }(5 \pi, \\
6 \pi) \cup[6 \pi, 7 \pi] \cup(7 \pi, 8 \pi) \text {. } \\
\because \quad \sin x=\frac{1}{3}, \\
\text { it is determined that } x \notin(5 \pi, 6 \pi), x \notin(7 \pi, 8 \pi) \text {. } \\
\text { Therefore, } x \in[6 \pi, 7 \pi] \text {. } \\
\text { Here } p=6 \pi \text {, and by formula (1) we get } \\
x=p+\frac{\pi}{2}+\arcsin \left[\sin \left(x-p-\frac{\pi}{2}\right)\right] \\
=6 \pi+\frac{\pi}{2}+\arcsin \\
{\left[\sin \left(x-6 \pi-\frac{\pi}{2}\right)\right]=6 \pi+\frac{\pi}{2}} \\
+\arcsin [-\cos x] \text { . } \\
\because \sin x=\frac{1}{3}, x \in(5 \pi, 8 \pi) \text {, } \\
\therefore \cos x=\frac{2 \sqrt{2}}{3} \text { or } \cos x=-\frac{2 \sqrt{2}}{3} \text {. } \\
\text { Thus, } x=6 \pi+\frac{\pi}{2}+\arcsin \frac{2 \sqrt{2}}{3} \\
=\frac{13 \pi}{2}+\arcsin \frac{2 \sqrt{2}}{3} \text {, } \\
\text { or } x=6 \pi+\frac{\pi}{2}+\arcsin \left(-\frac{2 \sqrt{2}}{3}\right) \\
=\frac{13 \pi}{2}-\arcsin \frac{2 \sqrt{2}}{3} \text {. } \\
\end{array}
$$

For the above formula, further research will reveal that appropriately refining the given interval can make the solutions to some problems more perfect.

In summary, it is extremely necessary to first refine the given interval when using the formula.