Trigonometric inequality - x domain 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 :)
 A: Since $\sqrt{(2\sqrt 2)^2+(\sqrt 2)^2}=\sqrt{10}$,
we have$$\begin{align}2\sqrt 2\sin x+\sqrt 2\cos x&=\sqrt{10}\left(\frac{2\sqrt{2}}{\sqrt{10}}\sin x+\frac{\sqrt 2}{\sqrt{10}}\cos x\right)\\&=\sqrt{10}\left(\frac{2}{\sqrt 5}\sin x+\frac{1}{\sqrt 5}\cos x\right)\\&=\sqrt{10}(\cos\theta\sin x+\sin \theta\cos x)\\&=\sqrt{10}\sin(x+\theta)\end{align}$$
where 
$$\cos\theta=\frac{2}{\sqrt 5},\quad \sin\theta=\frac{1}{\sqrt 5}\quad\Rightarrow \quad \tan\theta=\frac 12\quad\Rightarrow \quad \theta=\arctan(1/2)$$
Thus, for $k\in\mathbb Z$,
$$\begin{align}&2\sqrt 2\sin x+\sqrt 2\cos x\ge 0\\&\iff \sqrt{10}\sin(x+\arctan(1/2))\ge 0\\&\iff 0+2k\pi\le x+\arctan(1/2)\le \pi+2k\pi\\&\iff -\arctan(1/2)+2k\pi\le x\le \pi-\arctan(1/2)+2k\pi\end{align}$$
We can see that this is the same as
$$2\arctan(2-\sqrt 5)+2k\pi\le x\le 2\arctan(2+\sqrt 5)+2k\pi$$
because
$$\begin{align}&2\arctan(2-\sqrt 5)+\arctan(1/2)\\&=\arctan(2-\sqrt 5)+\arctan(2-\sqrt 5)+\arctan(1/2)\\&=\arctan(2-\sqrt 5)+\arctan\left(\frac{2-\sqrt 5+(1/2)}{1-(2-\sqrt 5)/2}\right)\\&=\arctan(2-\sqrt 5)+\arctan(\sqrt 5-2)\\&=0\end{align}$$
and
$$\begin{align}&2\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\pi+\arctan\left(\frac{2+\sqrt 5+(1/2)}{1-(2+\sqrt 5)/2}\right)\\&=\arctan(2+\sqrt 5)+\pi+\arctan(-2-\sqrt 5)\\&=\pi\end{align}$$
A: Clearly, $\sin x\cos x\ne0$
We first need $\sin x\cos x<0\implies\tan x<0$
$$2\sqrt2\sin x+\sqrt2\cos x=\sqrt{-2\sin x\cos x}$$
Divide both sides by $\sqrt{-2\sin x\cos x},$  to get $$2\sqrt{-\tan x}+\sqrt{-\cot x}=1$$
Let $\sqrt{-\tan x}=y>0,\tan x=-y^2,\sqrt{-\cot x}=\dfrac1y>0$
$$2y+\dfrac1y=1\iff2y^2-y+1=0\implies y=\dfrac{1\pm\sqrt{1-8}}4$$
But $y>0$
