\[\Leftrightarrow 2+cos2x+\sqrt{3}sin2x=3(sinx+\sqrt{3}cosx)\]
\[\Leftrightarrow 1+\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=3(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx)\]
\[\Leftrightarrow 1+sin(\frac{\pi}{6}+2x)=3sin(\frac{\pi}{3}+x)\]
Đặt \[t=\frac{\pi}{3}+x\]\[\Rightarrow \frac{\pi}{6}+2x=2t+\frac{\pi}{2}\]
pt trở thành: \[1+sin(2t+\frac{\pi}{2})=3sint\]
\[\Leftrightarrow 1-cos2t=3sint\]
\[\Leftrightarrow 2sin^{2}t-3sint=0\]
\[\Leftrightarrow sint=\frac{3}{2}\](loại) hoặc \[sint=0\]
\[\Leftrightarrow t=k\pi\]
\[\Rightarrow \frac{\pi}{3}+x=k\pi\]
\[\Leftrightarrow x=\frac{-\pi}{3}+k\pi\]