Example 3. Find the maximum and minimum values of the function $y=\sqrt[4]{7-2 \sin x}$ $+\sqrt[4]{2 \sin x+1}$.
Solution: The domain of the function is obviously $\mathrm{A}=\{\mathrm{x}: \left.\sin x \geqslant-\frac{1}{2}\right\}$, and it satisfies the conditions of the theorem.
$$
\text { Let } \begin{aligned}
\varphi(x) & =|7-2 \sin \mathrm{x}-(2 \sin \mathrm{x}+1)| \\
& =|6-4 \sin \mathrm{x}|, \mathrm{x} \in \mathrm{A} .
\end{aligned}
$$

Then when (1) $\sin x=-\frac{1}{2}$, $\varphi_{\text {maximum }}=8$,
$$
y_{\text {minimum }}=\sqrt[4]{8} .
$$
$$
\text { (2) } \begin{aligned}
\sin x & =1 \text {, } \varphi_{\text {minimum }}=2 , \\
y_{\text {maximum }} & =\sqrt[4]{5}+\sqrt[4]{3} .
\end{aligned}
$$