The number of solutions of $\sin 5\theta \cos 3\theta =\sin 9\theta \cos 7\theta$ in $\left[0,\frac{\pi }{2}\right]$ is

$\begin{array}{l}2\sin 5\theta \cos 3\theta =2\sin 9\theta \cos 7\theta \\ \Rightarrow \sin 8\theta +\sin 2\theta =\sin 16\theta +\sin 2\theta \\ \Rightarrow \sin 16\theta -\sin 8\theta =0\Rightarrow 2\cos 12\theta \sin 4\theta =0\\ \therefore \cos 12\theta =0\Rightarrow \theta =(2n+1)\frac{\pi }{24},\sin 4\theta =0\Rightarrow \theta =\frac{n\pi }{4}\\ \therefore \theta =\frac{\pi }{24},\frac{\pi }{8},\frac{5\pi }{24},\frac{7\pi }{24},\frac{3\pi }{8},\frac{11\pi }{24},0,\frac{\pi }{4},\frac{\pi }{2}\end{array}$