$2\sin^2x+\sin x-1= 0$
Đặt $t=\sin x \in [-1;1]$ ta có:
$2t^2+t-1=0$
$⇔ 2t^2+2t-t-1=0$
$⇔ 2t(t+1)-(t+1)=0$
$⇔ (2t-1)(t+1)=0$
$⇔ t=\dfrac12$ hoặc $t=-1$
$⇔ \sin x=\dfrac12$ hoặc $\sin x=-1$
$⇔ x=\dfrac{5\pi}6 + k2\pi$ hoặc $x=-\dfrac \pi 2+k2\pi \ (k\in \mathbb{Z})$
