Đáp án:
`cosx(sqrt3+sin2x)=cos2x (sinx+2)`
`<=>cosxsqrt3+cosx.2sinxcosx=sinxcos2x+2cos2x`
`<=>(cos2x-cosx).(2sinx+cosx)=0`
`<=>`\(\left[ \begin{array}{l}\cos2x=\cos x\\\sin x=-\dfrac{1}{2}\cos x\end{array} \right.\) `->` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{3\pi}{4}+k2\pi\end{array} \right.\) `(kinmathbbZ)`
