I5 (7-1, Yugoslavia) Find all real numbers $x$ in the interval $0 \leqslant x \leqslant 2 \pi$ that satisfy the inequality
$$
2 \cos x \leqslant|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leqslant \sqrt{2}
$$
$$
\begin{aligned}
& |\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \\
= & \left|\sqrt{(\sin x+\cos x)^{2}}-\sqrt{(\sin x-\cos x)^{2}}\right| \\
= & || \sin x+\cos x|-| \sin x-\cos x|| \\
= & \left\{\begin{array}{ll}
2|\cos x|, \quad \text { when } \frac{\pi}{4} \leqslant x \leqslant \frac{3}{4} \pi \text { or } \frac{5}{4} \pi \leqslant x \leqslant \frac{7}{4} \pi, \\
2|\sin x|, \quad \text { other cases. }
\end{array}\right.
\end{aligned}
$$

When $\frac{3}{4} \pi < x < \frac{5}{4} \pi$, $2|\cos x| > 0$.
The right inequality is obvious $(|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leqslant \sqrt{1+\sin 2 x} \leqslant \sqrt{2}$.
Therefore, the solution to the inequality is $\frac{\pi}{4} \leqslant x \leqslant \frac{7}{4} \pi$.