Find the general solution of each of the following equations:

To Find: General solution.

Given: $\sec ^{2} 2 x=1-\tan 2 x \Rightarrow 1+\tan ^{2} 2 x+\tan 2 x=1 \Rightarrow \tan 2 x(1+\tan 2 x)=0$

So, $\tan 2 x=0$ or $\tan 2 x=-1=\tan \left(\frac{3 \pi}{4}\right)$

Formula used: : $\tan \theta=0 \Longrightarrow \theta=\mathrm{n} \pi, \mathrm{n} \in \mid$ and $\tan \theta=\tan \alpha \Rightarrow \theta=\mathrm{k} \pi \pm \alpha, \mathrm{k} \in \mid$

By using above formula, we have

$2 x=n \pi$ or $2 x=k \pi \pm \frac{3 \pi}{4} \Rightarrow x=\frac{n \pi}{2}$ or $x=\frac{k \pi}{2} \pm \frac{3 \pi}{8}$

So the general solution is $x=\frac{n \pi}{2}$ or $x=\frac{k \pi}{2} \pm \frac{3 \pi}{8}$ where $n, k \in \mid$