\(\Leftrightarrow f\left(x\right)=sin^4x+cos^4x-2m.sinx.cosx\ge0;\forall x\)
Ta có: \(f\left(x\right)=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-2m.sinx.cosx\)
\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2-m\left(2sinx.cosx\right)\)
\(=-\frac{1}{2}sin^22x-m.sin2x+1\)
Đặt \(sin2x=t\Rightarrow f\left(t\right)=-\frac{1}{2}t^2-mt+1\ge0\) ; \(\forall t\in\left[-1;1\right]\)
\(a=-\frac{1}{2}< 0\) và \(ac=-\frac{1}{2}< 0\) nên bài toán thỏa mãn khi và chỉ khi:
\(t_1\le-1< 1\le t_2\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{2}f\left(-1\right)\le0\\-\frac{1}{2}f\left(1\right)\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}f\left(-1\right)\ge0\\f\left(1\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m+\frac{1}{2}\ge0\\\frac{1}{2}-m\ge0\end{matrix}\right.\) \(\Rightarrow-\frac{1}{2}\le m\le\frac{1}{2}\)
