已知函数f(x)=sin^2 x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4),x属于R,求f(x)的最小正周期和值域

f(x)=sin^2 x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)
=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)
=2-2cos2x+√3sin2x+(1/2)sin(2x-π/2)
=2+√3sin2x-3/2cos2x
=2+√3(1/2sin2x-√3/2cos2x)
=2+√3sin(2x-π/3)
由最小正周期公式得：T=2π/w=2π/2=π
f(x)(max)=2+√3
f(x)(min)=2-√3
所以f(x)的值域为【2-√3,2+√3】.