解：f(x)=2√3cos²x+2sinxcosx-m=2√3(1+cos2x)/2+sin2x-m=√3cos2x+sin2x+√3-m=2sin(2x+π/3)+√3-m2x+π/3=2kπ+π/2时，f(x)取最大值f(x)_max=2+√3-m=2∴m=√3f(x)的最小正周期：2π/2=π
