y'= m + cosx + $\frac{1}{2}cos2x +\frac{1}{3}cos3x$
= m+ cosx + $\frac{1}{2} (2cos^{2}x-1)+\frac{1}{3}(4cos^{3}x-3cosx)$
Đặt a= cosx
y tăng $\forall x\epsilon R$ <=> y' $\geq$ 0 $\forall x\epsilon R$
<=> m $\geq$ $\frac{-4}{3}a^{3}-a^{2}+\frac{1}{2} =g(a) \forall x\epsilon [-1;1]$
g'(a)=-4a2-2a(2a+1)=0 <=> a=$\frac{-1}{2} , a=0$
Lập Bảng biến thiên => Maxg(a) = g(-1)=$\frac{5}{6} \leq m$