$2\cos^2x-2\sin^2x=\sqrt{2}$
$⇔ \dfrac{\sqrt 2 }{2}\sin x - \dfrac{\sqrt 2 }{2}\cos x = \dfrac{1}{2}$
$⇔ \sin x\cos \dfrac{\pi}{4}-\cos x \sin \dfrac{\pi}{4}=\dfrac12$
$⇔ \sin x \left(x-\dfrac{\pi}{4}\right)=\dfrac12$
$⇔ \sin x \left(x-\dfrac{\pi}{4}\right)=\sin \dfrac{1}{6}$
\(⇔\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac \pi 6+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac {5\pi} {12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{array} \right. (k\in \mathbb{R})\)
Vậy $S=\left\{\dfrac {5\pi} {12}+k2\pi,\dfrac{13\pi}{12}+k2\pi \; \Bigg| \; k\in \mathbb R \right\}$
