Answer
$\frac{\pi}{8}+\frac{k\pi}{4}, \frac{\pi}{6}+2k\pi, \frac{5\pi}{6}+2k\pi$
Work Step by Step
Use the Sum-to-Product Formula $\sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$.
$\sin 5\theta-\sin 3\theta=\cos 4\theta$
$2\cos\frac{5\theta+3\theta}{2}\sin\frac{5\theta-3\theta}{2}=\cos 4\theta$
$2\cos 4\theta\sin \theta=\cos 4\theta$
$2\cos 4\theta\sin \theta-\cos 4\theta=0$
$\cos 4\theta(2\sin\theta-1)=0$
If $\cos 4\theta=0$, then $4\theta=\frac{\pi}{2}+k\pi$, and $\theta=\frac{\pi}{8}+\frac{k\pi}{4}$.
If $2\sin\theta-1=0$, then $2\sin \theta=1$, and $\sin \theta=\frac{1}{2}$. Then $\theta=\frac{\pi}{6}+2k\pi, \frac{5\pi}{6}+2k\pi$.
