Find $x$:
(1) $\sin x=\frac{1}{4}, x \in\left(\frac{\pi}{2}, \pi\right)$,
(2) $\sin x=-\frac{1}{3}, x \in\left(\pi, \frac{3}{2} \pi\right)$.
Analysis: (1) $\because \sin _{x=\frac{1}{4}}$,
$\therefore \arcsin \frac{1}{4} \in\left(0, \frac{\pi}{2}\right)$.
However, the problem is to find the angle within $\left(\frac{\pi}{2}, \pi\right)$ whose sine value is $\frac{1}{4}$. Using the induced formula, we get
$$
x=\pi-\arcsin \frac{1}{4}.
$$
(2) $\because \quad \sin x=-\frac{1}{3}$,
$\therefore \arcsin \left(-\frac{1}{3}\right) \in\left(-\frac{\pi}{2}, 0\right)$.
However, the problem is to find the angle within $\left(\pi, \frac{3}{2} \pi\right)$ whose sine value is $-\frac{1}{3}$. Using the unit circle and trigonometric
function lines (as shown in Figure 1),
$M^{\prime} M=N^{\prime} N$, it is
not difficult to derive
$$
x=\pi+\arcsin \frac{1}{3},
$$

or $x=\pi-\arcsin \left(-\frac{1}{3}\right)$.
When using inverse trigonometric functions to represent $x$ in $\sin x=k(|k|<1)$, if $x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$, then $x=\pi-\arcsin k$.