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!

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