(1)
f(x)=√3sinωx×cosωx-cos²ωx
=√3/2sin2wx-1/2(1+cos2wx)
=√3/2sin2wx-1/2cos2wx-1/2
=sin(2wx-π/6)-1/2
∵f(x) 的最小正周期为π
∴2π/(2w)=π,w=1
f(x)=sin(2x-π/6)-1/2
由-π/2+2kπ≤2x-π/6≤π/2+2kπ,k∈Z
得-π/6+kπ≤x≤π3+kπ,k∈Z
∴函数的单调递增区间为[-π/6+kπ,π/3+kπ],k∈Z
(2)
∵b²=ac
根据余弦定理：
cosx=(a²+c²-ac)/(2ac)≥(2ac-ac)/(2ac)=1/2=cosπ/3
∴0