\(\Leftrightarrow2\left(2cos^2x-1\right)-4\left(m+1\right)cosx+12m-22=0\)
\(\Leftrightarrow cos^2x-\left(m+1\right)cosx+3m-8=0\)
Đặt \(cosx=t\Rightarrow-1\le t\le1\)
\(\Rightarrow t^2-\left(m+1\right)t+3m-8=0\)
\(\Leftrightarrow t^2-t-8=m\left(t-3\right)\)
\(\Leftrightarrow m=\frac{t^2-t-8}{t-3}\)
Xét \(f\left(t\right)=\frac{t^2-t-8}{t-3}\) với \(t\in\left[-1;1\right]\)
\(f\left(t\right)-\frac{3}{2}=\frac{t^2-t-8}{t-3}-\frac{3}{2}=\frac{2t^2-5t-7}{2\left(t-3\right)}=\frac{\left(t+1\right)\left(7-2t\right)}{2\left(3-t\right)}\ge0\Rightarrow f\left(t\right)\ge\frac{3}{2}\)
\(f\left(t\right)-4=\frac{t^2-t-8}{t-3}-4=\frac{t^2-5t+4}{t-3}=\frac{\left(1-t\right)\left(t-4\right)}{3-t}\le0\Rightarrow f\left(t\right)\le4\)
\(\Rightarrow\frac{3}{2}\le f\left(t\right)\le4\Rightarrow\frac{3}{2}\le m\le4\)
