What is the General Solution of \(sin^2 \theta\) =\(sin^2 \alpha\) ?

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.

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