The general solution of the equation tan^2 x | vedclass

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(C) Given the equation $	an^2 x = 1$. Taking the square root on both sides,we get $	an x = \pm 1$. We know that $	an x = 	an \alpha$ implies $x =

Vedclass · Accepted Answer

$n \pi \pm \frac{\pi}{4}$

Vedclass · Answer

(C) Given the equation $	an^2 x = 1$. Taking the square root on both sides,we get $	an x = \pm 1$. We know that $	an x = 	an \alpha$ implies $x = n \pi + \alpha$. For $	an x = 1$,$x = n \pi + \frac{\pi}{4}$. For $	an x = -1$,$x = n \pi - \frac{\pi}{4}$. Combining these two results,the general solution is $x = n \pi \pm \frac{\pi}{4}$,where $n \in \mathbb{Z}$.

The general solution of the equation $\tan^2 x = 1$ is

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