\(\sin^2\theta – \sin \theta = 0\)
\(\sin \theta ( \sin \theta – 1 ) = 0\)
So either \(\sin \theta = 0\) or \(\sin \theta = 1\).
\(\sin \theta = 0\) for every multiple of \(\pi\). \(\sin A = 1\) for \(\frac{\pi}{2}\) plus every even multiple of \(\pi\).
So the solution is \(\theta = n\pi\) or \(\theta=\frac{\pi}{2} + 2 n\pi\).