Đáp án: $x\in\{-\dfrac{\pi+3k\pi}{6},\dfrac{\pi-6k\pi}{18}\}$
Giải thích các bước giải:
Ta có:
$\sqrt{3}\cos5x-2\sin3x\cos2x-\sin x=0$
$\to \sqrt{3}\cos5x-(\sin(3x+2x)+\sin(3x-2x))-\sin x=0$
$\to \sqrt{3}\cos5x-(\sin(5x)+\sin(x))-\sin x=0$
$\to \sqrt{3}\cos5x-\sin(5x)-\sin(x)-\sin x=0$
$\to \sqrt{3}\cos5x-\sin(5x)-2\sin(x)=0$
$\to \sqrt{3}\cos5x-\sin(5x)=2\sin(x)$
$\to \dfrac{\sqrt{3}}{2}\cdot\cos(5x)-\dfrac12\sin(5x)=\sin x$
$\to \sin(\dfrac{\pi}{3}-5x)=\sin x$
$\to \dfrac{\pi}{3}-5x=x+k2\pi\to x=\dfrac{\pi-6k\pi}{18}$
Hoặc $\dfrac{\pi}{3}-5x=\pi-x+k2\pi\to x=-\dfrac{\pi+3k\pi}{6}$
