Difficult trigonometric equation

Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$: $$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$

This is what I did: $$\begin{array}{crcl} \Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^2 x \\ \Longrightarrow &3\left(1-\sin^2 x\right)-\sin 2x &=& \sin^2 x \\ \Longrightarrow & 4\sin^2 x + \sin 2x - 3 &=& 0 \\ \Longrightarrow &2\left(1-\cos 2x\right)+\sin 2x - 3 &=& 0\\ \Longrightarrow &-2\cos 2x + \sin 2x &=& 1\end{array}$$

So, what's next?! Thank you in advance!

-
Just found : math.stackexchange.com/questions/656036/… –  lab bhattacharjee Jan 29 '14 at 18:12

HINT:

We have $$1-\sin2x+2\cos2x=0$$

Using Double-Angle Formulas, $$1-\frac{2t}{1+t^2}+2\frac{1-t^2}{1+t^2}=0$$ where $t=\tan x$

Solve the Quadratic Equation in $\tan x$

As $x\in\left[\frac{3\pi}2,2\pi\right], t=\tan x<0$

-
Thank you , I will try now! –  John G. Jan 29 '14 at 18:22
@user123499, yes please revert if you have any confusion –  lab bhattacharjee Jan 29 '14 at 18:25

Let $\displaystyle-2=r\cos\phi,1=r\sin\phi$ where $r>0$ so $\displaystyle\frac\pi2<\phi<\pi$ and $\displaystyle\tan\phi=-\frac12$

So we have $$r\cos(2x-\phi)=r\sin\phi$$

$$\implies \cos(2x-\phi)=\cos\left(\frac\pi2-\phi\right)$$

$$\implies 2x-\phi=2n\pi\pm\left(\frac\pi2-\phi\right)$$ where $n$ is any integer

Considering '+' sign, $\displaystyle2x-\phi=2n\pi+\left(\frac\pi2-\phi\right)\implies 2x=2n\pi+\frac\pi2$

Considering '-' sign, $\displaystyle2x-\phi=2n\pi-\left(\frac\pi2-\phi\right)=2n\pi-\frac\pi2+\phi\implies 2x=2n\pi+2\phi-\frac\pi2$

-
But can you tell me please, why do we have 2x−ϕ=2nπ±(π/2−ϕ) but not simply 2x−ϕ=π/2−ϕ? –  John G. Jan 29 '14 at 18:07
@user123499, as $\cos(-A)=\cos A$ and $\cos(2m\pi+y)=\cos y$ –  lab bhattacharjee Jan 29 '14 at 18:11
I know this , but I don't understand why do we write 2x−ϕ=2nπ±(π/2−ϕ) with 2πk? –  John G. Jan 29 '14 at 18:15
@user123499, as we have to find $x$ somewhere outside $[0,\frac\pi2]$ –  lab bhattacharjee Jan 29 '14 at 18:17
Thanks a lot!!! –  John G. Jan 29 '14 at 18:24

Say $$-2(\cos^2x-\sin^2x)=1-2\sin x\cos x=(\sin x-\cos x)^2$$ hence $x=k\pi+\frac{\pi}{4}$ and $x=k\pi+\pi+\frac{\pi}{4}$ also $\tan x=-\frac13$. So answer $\tan x=-\frac13$ is accept only.

-

from your last step 1−sin2x+2cos2x=0 =>(sin^2x+cos^2x)-2sinxcosx+2(cos^x-sin^2x)=0 =>3cos^2x-2sinxcosx-sin^2x=0 =>3cos^2x-3sinxcosx+sinxcosx-sin^2x=0 =>3cosx(cosx-sinx)+sinx(cosx-sinx)=0 =>(cosx-sinx)(3cosx+sinx)=0 =>cosx-sinx=0 or, 3cosx+sinx=0 =>tanx=1, or tanx=-3 =>x=π/4, or x=2π/3

-