Helppppppppppppppp Câu a thôi

$⇔1-2(\sin{x})^2+5\sin x+2=0$
$⇔2(\sin x)^2-5\sin x-3=0$
$⇔$ \(\left[ \begin{array}{l}\sin x=3 (loại)\\\sin x=\dfrac{-1}{2} \end{array} \right.\) 
$⇔$\(\left[ \begin{array}{l}x=\dfrac{- \pi}{6}+2k.\pi\\x=\dfrac{7\pi}{6} +2k. \pi\end{array} \right.\)