Ta có: \(\sin2x-cosx=\sqrt3\left(\sin x+cos2x\right)\)
=>\(\sin2x-\sqrt3\cdot cos2x=\sqrt3\cdot\sin x+cosx\)
=>\(\frac12\cdot\sin2x-\frac{\sqrt3}{2}\cdot cos2x=\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx\)
=>\(\sin\left(2x-\frac{\pi}{3}\right)=\sin\left(x+\frac{\pi}{6}\right)\)
=>\(\left[\begin{array}{l}2x-\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\ 2x-\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi=-x+\frac56\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}2x-x=\frac{\pi}{6}+\frac{\pi}{3}+k2\pi=\frac{\pi}{2}+k2\pi\\ 2x+x=\frac56\pi+\frac{\pi}{3}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{2}+k2\pi\\ x=\frac{7}{18}\pi+\frac{k2\pi}{3}\end{array}\right.\)
