已知函数f(x)=根号3sin(2x-π/6）+2sin的2次方（x-π/12）+sinπ/12cos5π/12,求使函

f(x)=√3sin(2x-π/6)-2sin^2(x-π/12)
根据公式cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α) 得：
f(x)=√3sin(2x-π/6)+cos(2x-π/6)-1
=2{(√3/2)sin(2x-π/6)+(1/2)cos(2x-π/6)}-1
=2{cos(π/6)sin(2x-π/6)+sin(π/6)cos(2x-π/6)}-1
再根据公式sin(α+β)=sinαcosβ+cosαsinβ得：
f(x)=2sin(2x)-1
所以f(x)的最小正周期为π,值域为[-3,1]