Example 1 Find the minimum value of $\frac{\sin x}{2}+\frac{2}{\sin x}(x \in(0, \pi))$.
Solution: Since the derivative of the function $f(t)=\frac{t}{2}+\frac{2}{t}(t \in(0,1])$ is
$$
f^{\prime}(t)=\frac{1}{2}-\frac{2}{t^{2}}=\frac{t^{2}-4}{2 t^{2}}<0,
$$

thus, $f(t)$ is strictly decreasing.
Therefore, the minimum value of $f(t)$ is $f(1)=\frac{5}{2}$.
Hence, the minimum value of $\frac{\sin x}{2}+\frac{2}{\sin x}$ is $\frac{5}{2}$, and it is achieved when $x=\frac{\pi}{2}$.