f(x)=√3sin(2x-π/6) + 2sin²(x-π/12)
=√3sin(2x-π/6)+1-cos(2x-π/6)
=2sin[(2x-π/6)-π/6] + 1
=2sin(2x-π/3) + 1
使函數(shù)取得最大值即 2x-π/3 = 2kπ+π/2
x ={x| kπ+5π/12,k∈Z}