已知函数f（x）=sin（2x+π/6）+sin（2x-π/6）+2cos²x（x∈R） ①求函数

f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos^2x
=(√3/2)sin(2x)+(1/2)cos(2x)+(√3/2)sin(2x)-(1/2)cos(2x)+1+cos(2x)
=√3sin(2x)+cos(2x)+1
=2[(√3/2)sin(2x)+(1/2)cos(2x)]+1
=2sin(2x+π/6)+1
f(x)max=3,最小正周期是π
f(x)≥2,即2sin(2x+π/6)≥1,即sin(2x+π/6)≥1/2
2kπ+π/6≤2x+π/6≤2kπ+(5/6)π
得kπ≤x≤kπ+π/3,k∈Z