The number of solutions of the equation sin5$\theta$ cos3$\theta$ = sin6$\theta$ cos2$\theta$ in [0,$\pi$] is

$$\begin{gathered} 2\sin 5\mathrm{\theta } \cos 3\mathrm{\theta }=2\sin 6\mathrm{\theta } \cos 2\mathrm{\theta } \\ \sin 8\mathrm{\theta }+\sin 2\mathrm{\theta }=\sin 8\mathrm{\theta }+\sin 4\mathrm{\theta } \\ \sin 4\mathrm{\theta }-\sin 2\mathrm{\theta }=0 \\ 2\mathrm{cos\theta } \mathrm{sin\theta }=0 \\ \cos 3\mathrm{\theta }=0 \mathrm{sin\theta }=0 \end{gathered}$$

$$\begin{gathered} \cos 3\mathrm{\theta }=0 \\ 3\mathrm{\theta }=\frac{\mathrm{\pi }}{2}, \frac{3\mathrm{\pi }}{2}, \frac{5\mathrm{\pi }}{2} \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{6}, \frac{3\mathrm{\pi }}{6}, \frac{5\mathrm{\pi }}{6} \\ \mathrm{sin\theta }=0 \\ \mathrm{\theta }=0, \mathrm{\pi } \\ \mathrm{no}. \mathrm{of} \mathrm{solves} 5 \end{gathered}$$