Given trigonometric equation is $\sin \left(9x\right)+\sin \left(3x\right)=0$

$\Rightarrow 2\sin 6x\cos 3x=0$

$\therefore$ either $\sin 6x=0$

or $\cos 3x=0$

for $x\in \left[0,2\pi \right]$

$\sin 6x=0$

$\Rightarrow x=0, \frac{\pi }{6}, \frac{\pi }{3}, \frac{\pi }{2}, \frac{2\pi }{3}, \frac{5\pi }{6}, \pi , \frac{7\pi }{6}, \frac{4\pi }{3}, \frac{3\pi }{2}, \frac{5\pi }{3}, 2\pi$

and $3x=0$

$\Rightarrow x=\frac{\pi }{6}, \frac{\pi }{2}, \frac{5\pi }{6}, \frac{7\pi }{6}, \frac{3\pi }{2}, \frac{11\pi }{6}$

So, number of solution of given equation is $13$.