已知函数f（x）=2cosxsin（x+兀/6）+cos^4x-sin^4x.求f（x）的最小正周期,若x属于[-兀/1

（1）f(x)=2cosx(sinx*√3/2+cosx*1/2)+(cos²x+sin²x)(cos²x-sin²x)
=√3cosxsinx+cos²x+cos²x-sin²x
=√3/2*sin2x+(1+cos2x)/2+cos²2x
=√3/2*sin2x+3/2*cos2x+1/2
=√3(sin2x*1/2+cos2x*√3/2)+1/2
=√3(sin2x*cosπ/3+cos2xsinπ/3)+1/2
=√3sin(2x+π/3)+1/2
T=2π/2=π
（2)∵x∈[-π/12,π/6]
∴2x+π/3∈[π/6,2π/3]
∴sin(2x+π/3)∈[1/2,1]
∴f(x)max=√3+1/2,此时x=π/12；
f(x)min=√3/2+1/2,此时x=-π/12