首先化简
f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx
=2cosxcos(x-π/6)-2sinx(√3/2sinx-1/2cosx)
=2cosxcos(x-π/6)-2sinx(sinxcosπ/6-cosxsinπ/6)
=2cosxcos(x-π/6)-2sinxsin(x-π/6)
=2cos(x+x-π/6)
=2cos(2x-π/6)
（1）f(x)的最小正周期T=2π/w=2π/2=π
（2）当x属于[-π/3,π/2]时,2x-π/6属于[-5π/6,5π/6]
所以cos(2x-π/6)属于[-√3/2,1]
所以f(x)的最小值为2*(-√3/2)=-√3