Example 1 Find the range of the function $y=\sqrt{x-4}+\sqrt{15-3 x}$.

Analysis: The general approach is: squaring, rearranging, isolating the radical, squaring again, and converting to a rational expression for solving. Since $4 \leqslant x \leqslant 5$, a trigonometric substitution can be used.
Solution: From $\left\{\begin{array}{l}x-4 \geqslant 0, \\ 15-3 x \geqslant 0,\end{array}\right.$ we get $4 \leqslant x \leqslant 5$.
Let $x=4+\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then
$y=\sin \theta+\sqrt{3} \cos \theta=2 \sin \left(\theta+\frac{\pi}{3}\right)$.
Since $\theta \in\left[0, \frac{\pi}{2}\right]$, therefore,
$\frac{\pi}{3} \leqslant \theta+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, when $\theta=\frac{\pi}{2}$, $y_{\text {min }}=1$;
when $\theta=\frac{\pi}{6}$, $y_{\text {max }}=2$.
Hence, the range of the function is $[1,2]$.
Comment: The function after the trigonometric substitution is concise and clear, but it must be particularly noted that the transformed function $y=2 \sin \left(\theta+\frac{\pi}{3}\right)$ is not a monotonic function on $\left[0, \frac{\pi}{2}\right]$, so the maximum value cannot be obtained at its left endpoint.