f[x]=根号3sin[2x-π ／6]+2sin²[x-π／12] [x∈R]

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

T=2π/2=π

使函数取得最大值即 2x-π/3 = 2kπ+π/2
    x ={x| kπ+5π/12,k∈Z}