f(x)=(√3sinwx-coswx)coswx+1/2
=2sin(wx-π/6)coswx+1/2
=sin(wx-π/6+wx)+sin(wx-π/6-wx)+1/2
=sin(2wx-π/6)-sinπ/6+1/2
=sin(2wx-π/6)
根據(jù)題意有：2π/2w=2π,所以w=1/2
f(x)=sin(x-π/6)
2bcosA=2c-√3a
2sinBcosA=2sinC-√3sinA
2sinBcosA=2sin(A+B)-√3sinA
2sinBcosA=2sinAcosB+2sinBcosA-√3sinA
2sinAcosB=√3sinA
因?yàn)閟inA≠0
所以cosB=√3/2
B=π/6
f（B）=sin(π/6-π/6）=0