$\frac{1}{2}$ sin x-$\frac{\sqrt[]{3} }{2}$ cosx=cos3x
sinx.sin$\frac{\pi}{6}$ -cosx.cos$\frac{\pi}{6}$=cos3x
cos(x+$\frac{\pi}{6}$)=-cos3x=cos(3x+ $\pi$ )
x+$\frac{\pi}{6}$ = 3x+$\pi$ +k2$\pi$ hoặc x+ $\frac{\pi}{6}$=-3x-$\pi$ +k2$\pi$
x=$\frac{-5\pi}{12}$ + k$\pi$ hoặc x=$\frac{-7\pi}{24}$+ $\frac{k\pi}{4}$