The number of solutions of the equation $\sin 5n\theta \cos 3n\theta =\sin 6n\theta .\cos 2n\theta$  in $\left[0,\pi \right]$  is

$\frac{2\sin 5n\theta \cos 3n\theta }{2}=\frac{2\sin 6n\theta .\cos 2n\theta }{2}$

$\sin 8n\theta +\sin 2n\theta =\sin 8n\theta +\sin 4n\theta$

$\sin 4n\theta -\sin 2n\theta =0$

$2\sin 2n\theta \cos 2n\theta -\sin 2n\theta =0$

$\sin 2n\theta \left(2\cos 2n\theta -1\right)=0$

$\sin 2n\theta =0\text{or}2\cos 2n\theta -1=0$

$2n\theta =k\pi$ $or2n\theta =2k\pi \pm \frac{\pi }{3}$

$\theta =\frac{k\pi }{2n}\theta =\frac{k\pi }{n}\pm \frac{\pi }{6n}$

$k\in z$           $k\in z$

No solutions lies between $\left[0,\pi \right]$  are 5