Note that $ \sqrt{1+sin2x}-\sqrt{1-sin2x}\leq\sqrt{2}$ is always true for all real x since $ \sqrt{1-sin2x}\geq\ 0$ and $ \sqrt{1+sin2z}\leq\sqrt{2}$ From $ 2cosx\leq\sqrt{2}$ $ \frac{\pi}{4}\leq\ x\leq\frac{7\pi}{4}$ Now consider $ 2cosx\leq\sqrt{1+sin2x}-\sqrt{1-sin2x}$ $ 4cos^{2}x\leq\ (1+sin2x)+(1-sin2x)-2\sqrt{1-sin^{2}2x}$ $ 4cos^{2}x\leq\ 2-2\sqrt{1-sin^{2}2x}$ $ \sqrt{1-sin^{2}2x}\leq\ 1-2cos^{2}x$ $ 1-sin^{2}2x\leq\ 1-4cos^{2}x+4cos^{4}x$ $ 1-4sin^{2}xcos^{2}x\leq\ 1-4cos^{2}x+4cos^{4}x$ $ 4cos^{4}x+4sin^{2}xcos^{2}x-4cos^{2}x\geq\ 0$ $ 4cos^{2}x(cos^{2}x+sin^{2}x-1)\geq\ 0$ But $ cos^{2}x+sin^{2}x-1=0$, so, $ 0\geq\ 0$ So it means that all x such that $ \frac{\pi}{4}\leq\ x\leq\frac{7\pi}{4}$ works

From $2\cos x\leq\sqrt{2}\iff\cos x\leq\tfrac{\sqrt{2}}{2}$, we find that $x\in[\tfrac{\pi}{4},\tfrac{7\pi}{4}]$ is necessary, and $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|\leq\sqrt{2}$ holds for all $x$, since the absolute difference between $\sqrt{1+\sin2x}$ and $\sqrt{1-\sin2x}$ attains its max. when $|\sin2x|$ attains its maximum of $1$, which yields $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|=\sqrt{2}$. Henceforth disregard the upper bound of $\sqrt{2}$. As for $2\cos x\leq|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$, assume that $2\cos x$ is nonnegative, because otherwise the inequality clearly holds since $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$ is nonnegative. By this assumption, both sides are nonnegative, so we can square both sides of our inequality to obtain $$4\cos^2x\leq(1+\sin2x)+(1-\sin2x)-2\sqrt{(1+\sin2x)(1-\sin2x)}=(1+1)-2\sqrt{1-\sin^22x}=2-2\cos2x=2-2(2\cos^2x-1)=4-4\cos^2x$$by the double-angle formula. Rearranging, $8\cos^2x\leq4\iff\cos x\leq\tfrac{\sqrt{2}}{2}$ (and $\geq0$). Combining this with the $\cos x<0$ solutions yields $\boxed{x\in[\tfrac{\pi}{4},\tfrac{7\pi}{4}]}$, which is precisely the interval from the $2\cos x\leq\sqrt{2}$ condition. Hence, this is the requested answer. $\blacksquare$

My work: $E=\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}=\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}-\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x}=$ $=\left| \sin x+\cos x \right|-\left| \sin x-\cos x \right|=\sqrt{2}\left( \left| \cos \left( \frac{\pi }{4}-x \right) \right|-\left| \sin \left( \frac{\pi }{4}-x \right) \right| \right)$ So ${{E}^{2}}=2\left( 1-\left| \sin \left( \frac{\pi }{4}-x \right)\cos \left( \frac{\pi }{4}-x \right) \right| \right)$ thus ${{E}^{2}}=2-\left| \sin \left( \frac{\pi }{2}-2x \right) \right|=2-\left| \cos 2x \right|\le 2\Rightarrow E\le \sqrt{2},\forall x\in \left[ 0,2\pi \right]$ On the other hand, we have $4{{\cos }^{2}}x\le 2-\left| \cos 2x \right|\Leftrightarrow 2\cos 2x\le -\left| \cos 2x \right|\Rightarrow \cos 2x<0$

Algebrical solution: Squaring all part, we get $4{{\cos }^{2}}x\le 2-2\sqrt{1-{{\sin }^{2}}2x}\le 2$ or $4{{\cos }^{2}}x\le 2-2\left| \cos 2x \right|\le 2$ and the rest is easy