Ta có $\sqrt{3}\cos 5x-2\sin 3x.\cos 2x-\sin x=0\iff \sqrt{3}\cos 5x-\sin 5x-\sin x-\sin x=0$

$\iff \sqrt{3}\cos 5x-\sin 5x=2\sin x\iff \frac{\sqrt{3}}{2}\cos 5x-\frac{1}{2}\sin 5x=\sin x\iff \sin \left(\frac{\pi }{3}-5x\right)=\sin x$

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

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