Question

Find the general solution of ${\sec }^{2}2x=1-\tan 2x$

Answer 1

Given ${\text{sec}}^{2}2x=1-\tan 2x$

$⟹1+{\tan }^{2}2x=1-\tan 2x$

$⟹\tan 2x=0,-1$

$2x=n\pi +0$ or $n\pi +\frac{3\pi }{4}$ where $n$ is an integer

$x=\frac{n\pi }{2}$ or $\frac{n\pi }{2}+\frac{3\pi }{8}$ where $n$ is an integer