Solution :
The general solution of \(sin \theta\) = \(sin \alpha\) is given by \(\theta\) = \(n\pi + (-1)^n \alpha\), n \(\in\) Z.
Proof :
We have, \(sin \theta\) = \(sin \alpha\)
\(\implies\) \(sin \theta\) – \(sin \alpha\) = 0
\(\implies\) \(2 sin ({\theta – \alpha\over 2}) cos({\theta + \alpha\over 2})\) = 0
\(\implies\) \(sin ({\theta – \alpha\over 2})\) = 0 or, \(cos({\theta + \alpha\over 2})\) = 0
\(\implies\) \({\theta – \alpha\over 2}\) = \(m\pi\) or \({\theta + \alpha\over 2}\) = \((2m + 1){\pi\over 2}\), m \(\in\) Z
\(\implies\) \(\theta\) = \(2m\pi + \alpha\), \(\in\) Z or, \(\theta\) = \((2m + 1)\pi – \alpha\), m \(\in\) Z.
\(\implies\) \(\theta\) = (any even multiple of \(\pi\)) + \(\alpha\) or, \(\theta\) = (any odd multiple of \(\pi\)) – \(\alpha\)
\(\implies\) \(\theta\) = \(n\pi + (-1)^n \alpha\), where n \(\in\) Z.
Remark : The equation \(cosec \theta\) = \(cosec \alpha\) is equivalent to \(sin \theta\) = \(sin \alpha\). Thus, \(cosec \theta\) = \(cosec \alpha\) and \(sin \theta\) = \(sin \alpha\) have the same general solution.
