<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] 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
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