Giải phương trình:  $\sqrt{3}\cos 5x-2\sin 3x\cos 2x-\sin x=0$.

 $\sqrt{3}\cos 5x-2\sin 3x\cos 2x-\sin x=0$

 $\iff \sqrt{3}\cos 5x-\left(\sin 5x+\sin x\right)-\sin x=0$

 $\iff \sqrt{3}\cos 5x-\sin 5x=2\sin x$

 $\iff \frac{\sqrt{3}}{2}\cos 5x-\frac{1}{2}\sin 5x=\sin x$

 $\iff \sin \frac{\pi }{3}.\cos 5x-\cos \frac{\pi }{3}.\sin 5x=\sin x$

 $\iff \sin \left(\frac{\pi }{3}-5x\right)=\sin x$

$\iff \left[\begin{array}{l}\frac{\pi }{3}-5x=x+k2\pi \\ \frac{\pi }{3}-5x=\pi -x+k2\pi \end{array}\right.\iff \left[\begin{array}{l}x=\frac{\pi }{18}+\frac{k\pi }{3}\\ x=-\frac{\pi }{6}+\frac{k\pi }{2}\end{array}\right.\left(k\in \mathbb{Z}\right)$

Vậy nghiệm của phương trình là:  $S=\left\{\frac{\pi }{18}+\frac{k\pi }{3};-\frac{\pi }{6}+\frac{k\pi }{2},k\in \mathbb{Z}\right\}$.