1. (YUG) Find all real numbers $x \in[0,2 \pi]$ such that
$$
2 \cos x \leq|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leq \sqrt{2} \text {. }
$$
1. Let us set $S=\mid \sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}$. Observe that $S^{2}=2-$ $2 \sqrt{1-\sin ^{2} 2 x}=2-2|\cos 2 x| \leq 2$, implying $S \leq \sqrt{2}$. Thus the righthand inequality holds for all $x$.
It remains to investigate the left-hand inequality. If $\pi / 2 \leq x \leq 3 \pi / 2$, then $\cos x \leq 0$ and the inequality trivially holds. Assume now that $\cos x>$ 0 . Then the inequality is equivalent to $2+2 \cos 2 x=4 \cos ^{2} x \leq S^{2}=$ $2-2|\cos 2 x|$, which is equivalent to $\cos 2 x \leq 0$, i.e., to $x \in[\pi / 4, \pi / 2] \cup$ $[3 \pi / 2,7 \pi / 4]$. Hence the solution set is $\pi / 4 \leq x \leq 7 \pi / 4$.