已知函数f(x)=sin^2(x-π/6)+sin^2(x+π/6)(1)求函数f(x)最小正周期(2)x∈[-π/3,

f(x)=(1/2)[1-cos(2x-π/3)]+(1/2)[1-cos(2x+π/3)]
=1-[cos(2x-π/3)+cos(2x+π/3)]
=1-2cos2xcos(π/3)
=1-cos2x
（1）周期为T=2π/2=π;
（2）f(x)在[-π/3,0]上是减函数,在[0,π/6]上是增函数,
最大值为f(-π/3)=3/2,最小值为f(0)=0