2. Find those $x \in[0,2 \pi]$, which satisfy the inequality

$$
2 \cos x \leq|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}|
$$
Solution. First, note that the inequality is meaningful for every value of $x$.

If $\cos x<0$, the left side of the inequality is negative, while the right side is non-negative, so the inequality is satisfied for all $x$ for which $\cos x<0$, i.e., for all $x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$.

Let $\cos x \geq 0$, i.e., $x \in\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$. In this case, both sides of the inequality are non-negative, so after squaring, the direction of the inequality changes, and we get

$$
\begin{aligned}
& 4 \cos ^{2} x \leq 2-2 \sqrt{\cos ^{2} 2 x} \\
& 2 \cos ^{2} 2 x \leq 1-|\cos 2 x| \\
& |\cos 2 x| \leq-\cos 2 x
\end{aligned}
$$

The last inequality is obviously satisfied for all $x$ for which $\cos 2 x \leq 0$, so taking into account that $x \in\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$, we get that $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, \frac{7 \pi}{4}\right]$.

Together with the other solutions (when $\cos x<0$), we get that $x \in\left[\frac{\pi}{4}, \frac{7 \pi}{4}\right]$.