AB·BC=|AB|*|BC|*cos
=accosθ=6
故：cosθ=6/(ac)
△ABC的面积：S=(1/2)acsinB
=(1/2)acsin(π-θ)=3tanθ
3≤S≤3√3
即：3≤3tanθ≤3√3
即：1≤tanθ≤√3
即：π/4≤θ≤π/3
2
f(θ)=sinθ^2+2sinθcosθ+3cosθ^2
=sin(2θ)+1+2cosθ^2
=sin(2θ)+1+(1+cos(2θ))
=sin(2θ)+cos(2θ)+2
=√2sin(2θ+π/4)+2
π/4≤θ≤π/3,即：π/2≤2θ≤2π/3
即：3π/4≤2θ+π/4≤11π/12
即：sin(2θ+π/4)∈[(√6-√2)/4,√2/2]
故f(θ)的最大值：√2*√2/2+3=3