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) 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

因为f(x)=2sin(wx-π/6)•sin(wx+π/3)，
所以f（x）=cos（wx-π/6-wx-π/3）-cos（wx-π/6+wx+π/3）=-cos（2wx+π/6）
因为最小正周期为π，所以w=1
所以f（x）=-cos（2x+π/6）
f(A)=f(B)=1/2，所以-cos（2A+π/6）=-cos（2B+π/6）=1/2
...

解:f(x)=2sin(wx-丌/6)sin(wx+兀/3)=2sin(wx-兀/6)cos(wx-兀/6)=Sin(2wx-兀/3).因为f(x)最小正周期为兀且W>O,所以2兀/2w=兀→w=|

f(x)=2sin(wx-π/6)•sin(wx+π/3)=2cos(wx+π/3)•sin(wx+π/3)=sin(2wx+2π/3)
由于周期为π，则w=1
f(x)=sin(2x+2π/3)
在三角形ABC中，若f(x)=sin(2x+2π/3)=1/2，则2x+2π/3=5π/6，或2x+2π/3=13π/6
解得x=π/12或x=3...