Đáp án+ Giải thích các bước giải:
$ \cos 2x-\sqrt{3}\sin 2x -\sqrt{3}\sin x-\cos x+4=0$
$ \Rightarrow \dfrac{1}{2}\cos 2x-\dfrac{\sqrt{3}}{2}\sin 2x-\dfrac{\sqrt{3}}{2}\sin x -\dfrac{1}{2}.\cos x +2=0$
$\Rightarrow \cos (2x+ \dfrac{\pi}{3})-\sin(x+\dfrac{\pi}{6})+2=0$
$\Rightarrow \cos 2(x+\dfrac{\pi}{6})-\sin(x+\dfrac{\pi}{6})+2=0$
$\Rightarrow 1-2.\sin^2 (x+\dfrac{\pi}{6})-\sin(x+\dfrac{\pi}{6})+2=0$
$\Rightarrow \left[ \begin{array}{l}\sin(x+\dfrac{\pi}{6})=1\\\sin(x+\dfrac{\pi}{6})=\dfrac{-3}{2}(L)\end{array} \right.$
$\Rightarrow x=\dfrac{\pi}{3}+k2\pi$
