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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 :)

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  • $\begingroup$ Did you find out when $sin2x$ is greater or equal to zero? $\endgroup$ – imranfat Mar 27 '16 at 18:53
  • $\begingroup$ Yes, I did, but I found out it's not enough, when I checked on Wolfram. $\endgroup$ – radzak Mar 27 '16 at 18:58
  • $\begingroup$ I think I did it. I just consider 2 instances. 1st when $$cosx>0$$ $$2tgx+1\ge0$$ $$tgx\ge-\frac{1}{2}$$ 2nd when $$cosx<0$$ $$2tgx+1\le0$$ $$tgx\le-\frac{1}{2}$$ $\endgroup$ – radzak Mar 27 '16 at 21:11
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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 $$\begin{align}&2\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\pi+\arctan\left(\frac{2+\sqrt 5+(1/2)}{1-(2+\sqrt 5)/2}\right)\\&=\arctan(2+\sqrt 5)+\pi+\arctan(-2-\sqrt 5)\\&=\pi\end{align}$$

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  • $\begingroup$ Wow, awesome, thanks!! I surely wouldn't come up with this on my own :D $\endgroup$ – radzak Mar 28 '16 at 9:33
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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 $y>0$

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