在ABC中,A,B,C为三个内角,f（B）=4cosB * sin^2（π/4 + B/2)+根号3 cos2B - 2cosB.

f(B)=4CosB*Sin^2(π/4 + B/2)+√3*Cos2B-2CosB
=4CosB(1-Cos[π/2 + B])/2+√3*Cos2B-2CosB
=Sin2B+√3*Cos2B
=2[Sin2B*(1/2)+Cos2B*(√3/2)]
=2Sin(2B+π/3)
若f(B)=2,则Sin(2B+π/3)=1,解得B=π/12
若f(B)-m>2h恒成立,则f(B)>2h+m
而f(B)的值域为[-1,1]
故2h+m