1
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jatimir 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 $$cosx<0$$ $$2tgx+1\le0$$ $$tgx\le-\frac{1}{2}$$ $\endgroup$ – Jatimir Mar 27 '16 at 21:11
1
$\begingroup$

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 $$\begin{align}&2\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\pi+\arctan\left(\frac{2+\sqrt 5+(1/2)}{1-(2+\sqrt 5)/2}\right)\\&=\arctan(2+\sqrt 5)+\pi+\arctan(-2-\sqrt 5)\\&=\pi\end{align}$$

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

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 $y>0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.