<s>[INST] 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!
 [/INST] 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 $x\in\left[\frac{3\pi}2,2\pi\right], t=\tan x<0$
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