Đáp án:
$S=\left\{\dfrac{\pi}{2}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin^2x-3\sin x+2=0$
$⇔\sin^2x-\sin x-2\sin x+2=0$
$⇔\sin x(\sin x-1)-2(\sin x-1)=0$
$⇔(\sin x-1)(\sin x-2)=0$
$⇔\left[ \begin{array}{l}\sin x-1=0\\\sin x-2=0\end{array} \right.⇔\left[ \begin{array}{l}\sin x=1\\\sin x=2\,(L)\end{array} \right.$
$⇔\sin x=1$
$⇔x=\dfrac{\pi}{2}+k2\pi\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{\pi}{2}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
