Question

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

Answer 1

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

$1+{\tan }^{2}2x=1-\tan 2x$

${\tan }^{2}2x+\tan 2x=0$

$\tan 2x(\tan 2x+1)=0$

$\tan 2x=0$ or $\tan 2x+1=0$

Now, $\tan 2x=0$

$\tan 2x=\tan 0$

$2x=n\pi +0,n\in Z$

$x=\frac{n\pi }{2},n\in Z$

$\tan 2x+1=0$

$\tan 2x=-1=-\tan \frac{\pi }{4}=\tan (\pi -\frac{\pi }{4})=\tan \frac{3\pi }{4}$

$2x=n\pi +\frac{3\pi }{4},n\in Z$

$x=\frac{n\pi }{2}+\frac{3\pi }{8},n\in Z$

Therefore, the general solution is $\frac{n\pi }{2}$ or $\frac{n\pi }{2}+\frac{3\pi }{8},n\in Z$