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