Đáp án:
$e)\left[\begin{array}{l} x=\dfrac{5\pi}{24}\pi+ \dfrac{k \pi}{2} (k \in \mathbb{Z}) \\ x=-\dfrac{7\pi}{12}+ k \pi (k \in \mathbb{Z})\end{array} \right.$
Giải thích các bước giải:
$e)\sin 3x-2\cos x=\sqrt{3} \cos 3x\\ \Leftrightarrow \sin 3x-\sqrt{3} \cos 3x=2\cos x\\ \Leftrightarrow \dfrac{1}{2}\sin 3x-\dfrac{\sqrt{3}}{2} \cos 3x=\cos x\\ \Leftrightarrow \sin \dfrac{\pi}{6}\sin 3x-\cos dfrac{\pi}{6}\cos 3x=\cos x\\ \Leftrightarrow \cos 3x\cos \dfrac{\pi}{6}-\sin 3x\sin \dfrac{\pi}{6}=-\cos x\\ \Leftrightarrow \cos \left(3x+\dfrac{\pi}{6}\right)=\cos (\pi-x)\\ \Leftrightarrow \left[\begin{array}{l} 3x+\dfrac{\pi}{6}=\pi-x+ k 2 \pi (k \in \mathbb{Z}) \\ 3x+\dfrac{\pi}{6}=x-\pi+ k 2 \pi (k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} 4x=\dfrac{5\pi}{6}\pi+ k 2 \pi (k \in \mathbb{Z}) \\ 2x=-\dfrac{7\pi}{6}+ k 2 \pi (k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{5\pi}{24}\pi+ \dfrac{k \pi}{2} (k \in \mathbb{Z}) \\ x=-\dfrac{7\pi}{12}+ k \pi (k \in \mathbb{Z})\end{array} \right.$
