题目内容：

已知函数f=cos(2x－π/3)－cos2x
1求函数f(x)的最小正周期

优质解答

f(x)=cos(2x－π/3)－cos2x
=cos2xcosπ/3+sin2xsinπ/3-cos2x
=√3/2*sin2x-1/2cos2x
=sin(2x-π/6)
T=2π/2=π
函数f(x)的最小正周期T=π