11. Let $0 \leqslant x \leqslant \pi$, and $3 \sin \frac{x}{2}=\sqrt{1+\sin x}-\sqrt{1-\sin x}$, then $\tan x=$
Solve
$$
3 \sin \frac{x}{2}=\left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|-\left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|=\left\{\begin{array}{l}
2 \sin \frac{x}{2}, 0 \leqslant x \leqslant \frac{\pi}{2} \\
2 \cos \frac{x}{2}, \frac{\pi}{2}<x \leqslant \pi
\end{array}\right.
$$

When $0 \leqslant x \leqslant \frac{\pi}{2}$, $\sin \frac{x}{2}=0 \Rightarrow \tan x=0$;
When $\frac{\pi}{2}<x \leqslant \pi$, $\tan \frac{x}{2}=\frac{2}{3}<1$, which does not meet the condition. In summary, $\tan x=0$.