\(y=m.sinx-\left(1+1-sin^2x\right)\)
\(y=sin^2x+m.sinx-2\)
Đặt \(sinx=t\Rightarrow\left|t\right|\le1\)
\(\Rightarrow y=t^2+mt-2\)
Để GTNN của hàm số hớn hơn -4 \(\Leftrightarrow t^2+mt-2>-4\) \(\forall t\in\left[-1;1\right]\)
\(\Leftrightarrow t^2+mt+2>0;\forall t\in\left[-1;1\right]\)
TH1: \(\Delta=m^2-8< 0\Leftrightarrow-2\sqrt{2}< m< 2\sqrt{2}\)
TH2: \(\left\{{}\begin{matrix}\Delta=0\\-\frac{b}{2a}\notin\left[-1;1\right]\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m=\pm2\sqrt{2}\\-\frac{m}{2}\notin\left[-1;1\right]\end{matrix}\right.\) \(\Rightarrow m=\pm2\sqrt{2}\)
TH3: \(m>2\sqrt{2}\Rightarrow-\frac{b}{2a}=-\frac{m}{2}< -1\Rightarrow f\left(t\right)_{min}>f\left(-1\right)=3-m\)
\(\Rightarrow3-m>0\Rightarrow m< 3\Rightarrow2\sqrt{2}< m< 3\)
TH4: \(m< -2\sqrt{2}\Rightarrow f\left(t\right)_{min}>f\left(1\right)=m+3>0\Rightarrow m>-3\)
\(\Rightarrow-3< m< -2\sqrt{2}\)
Vậy \(-3< m< 3\)
