Áp dụng công thức góc nhân đôi ta có $2sin^2{\frac{x}{2}}=1-cosx$ . Thay vào phương trình đã cho ta có
$2\sqrt{3}sinx(1+cosx)-2cosx(1-cosx)=3$
$\Leftrightarrow (2\sqrt3sinx-2cosx)+(2\sqrt3sinx.cosx+2cos^2x)=3$
$\Leftrightarrow (2\sqrt3sinx-2cosx)+(2\sqrt3sinx.cosx+cos2x+1)=3$
$\Leftrightarrow(2\sqrt3sinx-2cosx)+(\sqrt3sin2x+cos2x)=2 (1)$
Chú ý rằng
$2\sqrt3sinx-2cosx=4(\frac{\sqrt3}{2}sinx-\frac{1}{2}cosx)=4sin(x-\frac{\pi }{6})$
$\sqrt3sin2x+cos2x=2(\frac{\sqrt3}{2}sin2x+\frac{1}{2}cos2x)=2sin(2x+\frac{\pi}{6})$
Thay vào $(1)$ ta có
$2sin(x-\frac{\pi}{6})+sin (2x+\frac{\pi}{6})=1$
Đặt $x-\frac{\pi}{6}=a \rightarrow2x+\frac{\pi}{6}=2a+\frac{\pi}{2}$
$\Rightarrow2sina+sin(2a+\frac{\pi}{2})=1$
$\Rightarrow2sina+cos2a=1$
$\Rightarrow2sin a+1-2sin^2a=1$
Vậy $sina=0$ hoặc $sina=1$
Nếu $sina=0\Rightarrow sin(x-\frac{\pi}{6})=0\Rightarrow x=k\pi+\frac{\pi}{6}$
Nếu $sina=1\Rightarrow sin (x-\frac{\pi}{6})=1\Rightarrow x=2k\pi+\frac{\pi}{2}+\frac{\pi}{6}=2k\pi+\frac{2\pi}{3}$