f(x)=sin(x π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数）
应该是f(x)=sin(x +π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数）?
f(x)=sin(x+ π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数）
=2sinxcosπ/6+acosx+b
=√3sinx+acosx+b
=√(3+a^2)sin(x+θ)+b (sinθ=a/√(3+a^2),cosθ=√3/√(3+a^2),
f（x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2
∴-π/3+θ=-π/2,b-√(3+a^2)=2
θ=-π/6
a=-1,b=2+2=4