\(4\left(cosx+1\right)+4\left(1-cos^2x\right)-5-3m=0\)
\(\Leftrightarrow-4cos^2x+4cosx+3=3m\)
Đặt \(f\left(x\right)=-4cos^2x+4cosx+3\)
\(f\left(x\right)=-\left(2cosx-1\right)^2+4\le4\)
\(f\left(x\right)=-4cos^2x+4cosx+8-5=4\left(cosx+1\right)\left(2-cosx\right)-5\ge-5\)
\(\Rightarrow-5\le f\left(x\right)\le4\)
\(\Rightarrow-5\le3m\le4\Rightarrow-\frac{5}{3}\le m\le\frac{4}{3}\)