1. (YUG) Find all real numbers $x \in[0,2 \pi]$ such that  $$ 2 \cos x \leq|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leq \sqrt{2} $$
1. Let us set \( S = |\sqrt{1 + \sin 2x} - \sqrt{1 - \sin 2x}| \). Observe that \( S^2 = 2 - 2 \sqrt{1 - \sin^2 2x} = 2 - 2|\cos 2x| \leq 2 \), 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 \( 2 + 2 \cos 2x = 4 \cos^2 x \leq S^2 = 2 - 2|\cos 2x| \), which is equivalent to \( \cos 2x \leq 0 \), i.e., to \( x \in [\pi / 4, \pi / 2] \cup [3 \pi / 2, 7 \pi / 4] \). Hence the solution set is \( \pi / 4 \leq x \leq 7 \pi / 4 \).