The values of $\theta$ satisfying sin 5$\theta$ = sin 3$\theta$ - sin $\theta$ and 0 < $\theta$ < $\frac{\pi }{2}$

$$\begin{gathered} \sin 5\mathrm{\theta }=\sin 3\mathrm{\theta }-\mathrm{sin\theta } \\ \sin 3\mathrm{\theta }=\sin 5\mathrm{\theta }+\mathrm{sin\theta } \\ \sin 3\mathrm{\theta }=2\mathrm{sin\theta } \cos 2\mathrm{\theta } \\ 2\sin 3\mathrm{\theta } \cos 2\mathrm{\theta }-\sin 3\mathrm{\theta }=0 \\ \sin 3\mathrm{\theta }(2\cos 2\mathrm{\theta }-1)=0 \\ \sin 3\mathrm{\theta }=0, 2\cos 2\mathrm{\theta }=1 \\ 3\mathrm{\theta }=\mathrm{\pi } , \cos 2\mathrm{\theta }=\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3} \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{3}, 2\mathrm{\theta }=\frac{\mathrm{\pi }}{3}\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{6} \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{3}, \frac{\mathrm{\pi }}{6} \end{gathered}$$