f(x)=√3sin2x+2cos^2x+m=√3sin2x+cos2x+1+m=2sin（2x+π/6）+1+mx属于[0,π/2] ∴2x+π/6属于[π/6,7π/6] ∴sin（2x+π/6）最大值为1∴2+1+m=6m=3.∴f(x)=2sin（2x+π/6）+4f(x)≥5即sin（2x+π/6）≥1/2π/6+2k...