Đáp án:
Ta có:
$2sin2x+1=0$ $⇔2sin2x=-1$
$⇔sin2x=-0,5$
Vì: $sinx=$$\alpha=sin\alpha$
$⇔$\(\left[ \begin{array}{l}2x=-\pi/6+k2\pi\\2x=\pi+\pi/6+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\pi/12+k\pi\\x=7\pi/12+k\pi\end{array} \right.\)
Vậy $S=${$-\pi/12+k\pi; 7\pi/12+k\pi(k∈Z)$}
BẠN THAM KHẢO NHA!!!