<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] $\sqrt{3\cos^2(x)-\sin(2x)}=-\sin(x)$
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 $-\sin(x)$.
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