the following equality is given: $$2 \sqrt2\sin{}x+\sqrt2\cos{x}=\sqrt{-\sin2x}$$
I managed to solve it, but I have a problem with this inequality to define x domain:$$2 \sqrt2\sin{}x+\sqrt2\cos{x}\ge0$$
Wolfram says: $$x \in\left\langle 2\arctan(2- \sqrt{5}) + 2k \pi, 2\arctan(2+ \sqrt{5} ) + 2k \pi \right\rangle$$
but I don't know how to get to this point. I've tried some transformations and came up with this:
$$3\cos(\frac{\pi}{4}-x)-\cos(x+\frac{\pi}{4})\ge0$$
and still can't go any further.
Any help will be appreciated :)