Đáp án + Giải thích các bước giải:
$\sin^22x+\sin^2x=1$
$⇔ \dfrac{1-\cos 2x}{2}+1-\cos^22x=1$
$⇔ \cos^22x+\dfrac12 \cos 2x - \dfrac12 = 0$
\(\Leftrightarrow\left[ \begin{array}{l}\cos 2x=-1\\\cos 2x=\dfrac12\end{array} \right.\)
\(\Leftrightarrow\left[ \begin{array}{l}2x=\pi+k2\pi\\2x=\pm\dfrac{\pi}{3}+k2\pi\end{array} \right.\)
\(\Leftrightarrow\left[ \begin{array}{l}x=\dfrac \pi 2+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{array} \right. (k\in\mathbb{Z})\)
Vậy $S=\left\{\dfrac \pi 2+k\pi ;\pm\dfrac{\pi}{6}+k\pi\ \Big |\Big. \ k\in \mathbb{Z}\right\}$
