Find the general solution of the equation

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

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

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

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

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

Now, $\tan 2 \mathrm{x}=0$

$\Rightarrow \tan 2 \mathrm{x}=\tan 0$

$\Rightarrow 2 \mathrm{x}=\mathrm{n} \pi+0$, where $\mathrm{n} \in \mathrm{Z}$

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

$\tan 2 x+1=0$

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

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

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

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