\(\Leftrightarrow\sqrt{3}sin2x-sinx+2=\sqrt{3}cosx-cos2x\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x-\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx+1=0\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)-sin\left(x+\dfrac{\pi}{3}\right)+1=0\)
Đặt \(x+\dfrac{\pi}{3}=t\Rightarrow x=t-\dfrac{\pi}{3}\Rightarrow2x+\dfrac{\pi}{6}=2t-\dfrac{\pi}{2}\)
\(\Rightarrow sin\left(2t-\dfrac{\pi}{2}\right)-sint+1=0\)
\(\Leftrightarrow-cos2t-sint+1=0\)
\(\Leftrightarrow2sin^2t-1-sint+1=0\)
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