7. Find the maximum value of the function $f(x)=\sqrt{1+\sin x}+\sqrt{1-\sin x}+\sqrt{2+\sin x}+\sqrt{2-\sin x}+\sqrt{3+\sin x}+$ $\sqrt{3-\sin x}$.
7. Let $a_{1}=(\sqrt{1+\sin x}, \sqrt{1-\sin x}), a_{2}=(\sqrt{1-\sin x}, \sqrt{1+\sin x})$,
$$
\begin{array}{l}
a_{3}=(\sqrt{2+\sin x}, \sqrt{2-\sin x}), a_{4}=(\sqrt{2-\sin x}, \sqrt{2+\sin x}), \\
a_{5}=(\sqrt{3+\sin x}, \sqrt{3-\sin x}), a_{6}=(\sqrt{3-\sin x}, \sqrt{3+\sin x}),
\end{array}
$$

Then $a_{1}+a_{2}+a_{3}+a_{1}+a_{5}+a_{6}=(f(x), f(x))$,
Thus, $\left|a_{1}+a_{2}+\cdots+a_{6}\right|=f(x) \cdot \sqrt{2}$.
And $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{6}\right|=2(\sqrt{2}+2+\sqrt{6}) \geqslant f(x) \cdot \sqrt{2}$,
Therefore, $f(x) \leqslant 2+2 \sqrt{2}+2 \sqrt{3}$.
Equality holds if and only if $\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}=\frac{\sqrt{1-\sin x}}{\sqrt{1+\sin x}}=\frac{\sqrt{2+\sin x}}{\sqrt{2-\sin x}}=\frac{\sqrt{2-\sin x}}{\sqrt{2+\sin x}}=\frac{\sqrt{3+\sin x}}{\sqrt{3-\sin x}}=\frac{\sqrt{3-\sin x}}{\sqrt{3+\sin x}}$, and the numerators

have the same sign.
Hence $\sin x=0$, i.e., $x=k \pi, k \in \mathbf{Z}$, when
$$
f(x)_{\max }=2+2 \sqrt{2}+2 \sqrt{3}
$$