$9 \cdot 3$ Find all real numbers $x$ that satisfy the following conditions:
$$0 \leqslant x \leqslant 2 \pi \text { and } 2 \cos x \leqslant|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leqslant \sqrt{2} \text {. }$$
[Solution] Clearly, for any real number $x$ we have
$$|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leqslant \sqrt{2}$$

The problem reduces to solving the inequality in $[0,2 \pi]$
$$2 \cos x \leqslant|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}|$$

When $\frac{\pi}{2} \leqslant x \leqslant \frac{3 \pi}{2}$, since $\cos x \leqslant 0$, (1) holds.
When $0 \leqslant x < \frac{\pi}{2}$ or $\frac{3 \pi}{2} < x \leqslant 2\pi$, $\cos x > 0$ and
$$|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}|^{2}=2-2|\cos 2 x|$$

Thus, (1) is equivalent to
$$4 \cos ^{2} x \leqslant 2-2|\cos 2 x|$$

which simplifies to
$$|\cos 2 x| \leqslant-\cos 2 x$$

Hence, $\cos 2 x \leqslant 0$. Therefore, $\frac{\pi}{4} \leqslant x < \frac{\pi}{2}$ or $\frac{3 \pi}{2} < x \leqslant \frac{7 \pi}{4}$.
In summary, the solution to the original problem is $\frac{\pi}{4} \leqslant x \leqslant \frac{7 \pi}{4}$.