What is the General Solution of \(cos \theta\) = \(cos \alpha\) ?

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.

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