11. Let $0 \leqslant x \leqslant \pi$, and
$$
3 \sin \frac{x}{2}=\sqrt{1+\sin x}-\sqrt{1-\sin x} \text {. }
$$

Then $\tan x=$ . $\qquad$
11. 0 .
Notice that the given equation can be transformed into
$9 \sin ^{2} \frac{x}{2}=2-2 \sqrt{1-\sin ^{2} x}=2-2|\cos x|$.
If $0 \leqslant x \leqslant \frac{\pi}{2}$, then
$9 \sin ^{2} \frac{x}{2}=2-2 \cos x=4 \sin ^{2} \frac{x}{2}$
$\Rightarrow \sin \frac{x}{2}=0 \Rightarrow \tan x=0$;
If $\frac{\pi}{2}<x \leqslant \pi$, then
$9 \sin ^{2} \frac{x}{2}=2+2 \cos x=4 \cos ^{2} \frac{x}{2}$
$\Rightarrow \tan \frac{x}{2}=\frac{2}{3}<1$ (not valid, discard).
Therefore, $\tan x=0$.