Đáp án + Giải thích các bước giải:
$4\sin^2x=3$
$⇔ 4(1-\cos 2x)=6$
$⇔ 4-4\cos 2x=6$
$⇔ 4\cos 2x=-2$
$⇔ \cos 2x=\dfrac{-1}{2}$
\(⇔ \left[ \begin{array}{l}2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
\(⇔ \left[ \begin{array}{l}x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right. (k\in \mathbb{Z})\)
Vậy $S=\left\{\dfrac{\pi}{3}+k\pi;-\dfrac{\pi}{3}+k\pi\ \Big|\Big. k\in \mathbb{Z}\ \right\}$
