Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ 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!
 A: 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, $\displaystyle2x-\phi=2n\pi-\left(\frac\pi2-\phi\right)=2n\pi-\frac\pi2+\phi\implies 2x=2n\pi+2\phi-\frac\pi2$
A: 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 $x\in\left[\frac{3\pi}2,2\pi\right], t=\tan x<0$
A: 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
A: $\sqrt{3\cos^2(x)-\sin(2x)}=-\sin(x)$
You square the both sides
$3\cos^2(x)-\sin(2x)=\sin^2(x)$
By using double angle identity of $\sin$ and dividing all the terms by $\cos^2(x)$
$\tan^2(x)+2\tan(x)-3=0$
Quadratic equation 
$\tan(x)=1\text{ or }-3$
Find all the solutions in the interval 
$x=5.03\text{rad}$ 
don't forget to check this value in the original equation.
Here when you substitute this value in the equation it equals zero but if the interval is $[0,2\pi]$, one of the solutions is $\frac{\pi}{4}$ but it doesn't work, it works if the right side is $\sin(x)$ not $-\sin(x)$.
