f(x)=cos(2x-π/3)+sin²x-cos²x=cos2xcos(π/3)+sin2xsin(π/3)-cos2x=cos2x*(1/2)+sin2x*(√3/2)-cos2x=(√3/2)*sin2x-(1/2)*cos2x=sin(2x-π/6)所以函数f(x)的最小正周期是T=2π/2=π令2x-π/6=kπ+π/2...