PT $\Leftrightarrow -[2\sin^2 x + \sin x -(1+m)] = 0$
$\Leftrightarrow 2\sin^2 x +\sin x -(1+m) =0$
$\Delta = 9 + 8m$
$\Rightarrow$ PT có nghiệm $\forall m > \frac{-9}{8}$
$\Rightarrow$ nghiệm $ \sin x = \frac{-1\pm \sqrt{9+8m}}{4}$
để $\sin x$ có nghiệm $\in [0;\pi]$
$\Rightarrow 0 \leq \sin x \leq1 \Rightarrow 0\leq \frac{-1\pm \sqrt{9+8m}}{4}\leq1 $
$\Leftrightarrow 1 \leq \pm \sqrt{9+8m} \leq 5$