Solution :
The general solution of \(sin^2 \theta\) = \(sin^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z.
Proof :
We have, \(sin^2 \theta\) =\(sin^2 \alpha\)
\(\implies\) \(2 sin^2 \theta\) =\(2 sin^2 \alpha\)
\(\implies\) \(1 – cos 2\theta\) = \(1 – cos 2\alpha\)
\(\implies\) \(cos 2\theta\) = \(cos 2\alpha\)
\(\implies\) \(2\theta\) = \(2n\pi \pm 2\alpha\), n \(\in\) Z.
\(\implies\) \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z.
