f(x) =cos2xcosπ/3+sin2xsinπ/3-2*(1-cos2x)/2=√3(√3/2*cos2x+1/2*sin2x)-1=√3sin(2x+π/3)-1
故其最小正周期为2π/2=π.
增区间：[kπ-5π/12,kπ+π/12],k∈Z
减区间：[kπ+π/12,kπ+7π/12],k∈Z