$\sec^22x=1-\tan 2x$
$1+\tan^22x=1-\tan 2x$
$\tan^22x+\tan 2x=0$
$\tan 2x(1+\tan 2x)=0$
$\tan 2x=0$ or $\tan 2x+1=0$
$\Rightarrow \tan 2x=0$ or $\tan 2x=-1$
Now $\tan 2x=0\Rightarrow 2x=n\pi$
$x=\large\frac{n\pi}{2}$$,n\in I$
Again $\tan 2x=-1\Rightarrow \tan 2x=\tan(-\large\frac{\pi}{4})$
$2x=n\pi-\large\frac{\pi}{4}$
$x=\large\frac{n\pi}{2}-\frac{\pi}{8}$$\quad n\in I$
Hence (A) is the correct answer.