$\sin^2x+\sin 2x+2\cos^2x=2$
$⇔ 2\cos x\sin x - \sin^2x+ 2\cos^2x = 2(\sin^2x+\cos^2x)$
$⇔ 2\cos x\sin x - \sin^2x = 0$
$⇔ (2\cos x-\sin x)\sin x=0$
\(⇔ \left[ \begin{array}{l}2\cos x-\sin x=0\\\sin x=0\end{array} \right.\)
\(⇔ \left[ \begin{array}{l}2\cos x=\sin x\\\sin x=0\end{array} \right.\)
\(⇔ \left[ \begin{array}{l}x= \arctan (2)+k\pi\\ x=k\pi\end{array} \right. (k\in \mathbb{Z})\)
Vậy $S=\left\{\arctan (2)+k, k\pi ; k\in \mathbb{Z}\right\} $
