已知函数f(x)=2sin(wx-π/6)•sin(wx+π/3)(其中w>0,x∈R的最小正周期为π).问：（1）求W的值.（2

(1) f(x)=2sin(wx-π/6)•sin(wx+π/2-π/6)
=2sin[π/2+(wx-π/6)]•sin(wx-π/6)
=2cos(wx-π/6)•sin(wx-π/6)
=sin(2wx-π/3)
因周期T=2π/2w=π,则w=1
所以f(x)=sin(2x-π/3)
在三角形ABC中,若A〈B,且f(A)=f(B)=1/2
则由f(x)=sin(2x-π/3)=1/2,
知2A-π/3=π/6,A=π/4
2x-π/3=π-π/6,B=7π/12
所以C=π-A-B=π-π/4-7π/12=π/6
由正弦定理BC/AB=sinA/sinC
=sin(π/4)/sin(π/6)
=(√2/2)/(1/2)
=√2