F(x) = √3sin(2x-π/6)-cos(2x-π/6)= 2 { sin(2x-π/6)*cosπ/6 - cos(2x-π/6)sinπ/6 }= 2 sin(2x-π/6-π/6)= 2 sin(2x-π/3)最小正周期=2π/2 = π2x-π/3 = 2kπ+π/2时函数有最大值此时：x∈kπ+5π/12,其中...