$\sqrt{2}\sec x+\tan x=1$

We have,

$\sqrt{2}\sec x+\tan x=1$

 $\frac{\sqrt{2}}{\cos x}+\frac{\mathrm{sinx}}{\mathrm{cosx}}=1$

 $\sqrt{2}+\sin x=\cos x$

 $\cos x-\sin x=\sqrt{2}$

 $\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x=1$

 $\cos \frac{\pi }{4}\cos x-\sin \frac{\pi }{4}\sin x=1$

 $\cos \left(x+\frac{\pi }{4}\right)=\cos 0$

 $x+\frac{\pi }{4}=2n\pi$

 $x=2n\pi -\frac{\pi }{4}$

Hence, this is the value x is $2n\pi -\frac{\pi }{4}$.