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

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

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

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

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

Now, $\tan 2 x=0$

$\Rightarrow \tan 2 x=\tan 0$

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

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

$\tan 2 x+1=0$

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

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

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

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