Đáp án: $x=\dfrac{\pi}{42}+k\dfrac{2\pi}{7}$ và $x=\dfrac{-\pi}{6}+k2\pi$ ($k\in\mathbb Z$)
Giải thích các bước giải:
$\Rightarrow \sin x+\cos x\sin 2x+\sqrt3\cos 3x=2(\cos4x+\sin 3x)$
$\Rightarrow \sin x+2{\cos}^2x\sin x-2{\sin}^3x+\sqrt3\cos 3x=2\cos 4x$
$\Rightarrow \sin x+2\sin x({\cos}^2x-{\sin}^2x)+\sqrt3\cos 3x=2\cos 4x$
$\Rightarrow \sin x+2\sin x\cos 2x+\sqrt3\cos 3x=2\cos 4x$
$\Rightarrow \sin x+2.\dfrac{1}{2}[\sin 3x+\sin(-x)]+\sqrt3\cos 3x=2\cos 4x$
$\Rightarrow \sin 3x+\sqrt3\cos 3x=2\cos 4x$
$\Rightarrow \dfrac{1}{2}\sin 3x+\dfrac{\sqrt3}{2}\cos 3x=\cos 4x$
$\Rightarrow \sin(3x+\dfrac{\pi}{3})=\sin(\dfrac{\pi}{2}-4x)$
$\Rightarrow 3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-4x+k2\pi$ hoặc $3x+\dfrac{\pi}{3}=\pi-(\dfrac{\pi}{2}-4x)+k2\pi$
Vậy $x=\dfrac{\pi}{42}+k\dfrac{2\pi}{7}$ và $x=\dfrac{-\pi}{6}+k2\pi$ ($k\in\mathbb Z$).
