Answer
$x=\dfrac{5\pi}{4}, \dfrac{ 7\pi}{4}$
Work Step by Step
We have $ \cos (x +\dfrac{\pi}{4}) =\cos x \cos \dfrac{\pi}{4}-\sin x \sin \dfrac{\pi}{4}=\dfrac{\sqrt 2}{2} \cos x -\dfrac{\sqrt 2}{2} \sin x$
$ \cos (x -\dfrac{\pi}{4}) =\cos x \cos \dfrac{\pi}{4}+\sin x \sin \dfrac{\pi}{4}=\dfrac{\sqrt 2}{2} \cos x +\dfrac{\sqrt 2}{2} \sin x$
After simplifying, we get $\dfrac{\sqrt 2}{2} \cos x -\dfrac{\sqrt 2}{2} \sin x - \dfrac{\sqrt 2}{2} \cos x -\dfrac{\sqrt 2}{2} \sin x= -\sqrt 2 \sin x=1$
or, $\sin x =\dfrac{-1}{\sqrt 2}$
or, $\sin x=\dfrac{-\sqrt 2}{2}$
This yields: $x=\dfrac{5\pi}{4}, \dfrac{ 7\pi}{4}$
