求函数f(x)=√3sin(2x-π／6)+2平方sin(x-π／12)的最小正周期及取得最大值x的集合?

f(x)=√3sin(2x-π／6)+2平方sin(x-π／12)
=√3sin(2x-π／6)+1-cos(2x-π／6)
=2(√3/2sin(2x-π／6)-1/2cos(2x-π／6))+1
=2(sin(2x-π／6)cosπ／6-cos(2x-π／6)sinπ／6)+1
=2sin(2x-π／6-π／6)+1
=2sin(2x-π/3)+1
所以最小正周期T=2π/2=π
当sin(2x-π/3)=1时取得最大值
2x-π/3=2kπ+π/2
2x=2kπ+5π/6
x=kπ+5π/12

f(x)=√3sin(2x-π／6)+2sin²(x-π／12)=√3sin(2x-π／6)-cos(2x-π／6)+1=2sin（2x-π／3）+1.
故T=π，由2x-π／3=2kπ+π/2得，x=kπ+5π/12，k∈Z，
函数f(x)取得最大值x的集合{x|x=x=kπ+5π/12,k∈Z}