Hàm xác định trên R khi và chỉ khi:
\(sin^2x-m.sinx+1>0\) ; \(\forall x\in R\)
Đặt \(sinx=t\in\left[-1;1\right]\)
\(\Rightarrow f\left(t\right)=t^2-m.t+1>0\) ;\(\forall t\in\left[-1;1\right]\)
\(\Leftrightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)>0\)
TH1: \(-\frac{b}{2a}=\frac{m}{2}\in\left[-1;1\right]\Rightarrow-2\le m\le2\)
\(\Rightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)=f\left(\frac{m}{2}\right)=\frac{4-m^2}{4}>0\Rightarrow-2< m< 2\)
TH2: \(\frac{m}{2}>1\Leftrightarrow m>2\Rightarrow f\left(t\right)_{min}=f\left(1\right)=2-m>0\Rightarrow m< 2\left(ktm\right)\)
TH2: \(m< -2\Rightarrow f\left(t\right)_{min}=f\left(-1\right)=m+2>0\Rightarrow m+-2\left(ktm\right)\)
Vậy \(-2< m< 2\)
