在△ABC中,角A,B,C所对的边分别为a,b,c,m=(cos(x-B),cosB),n=(cosb,-1/2）求角B

m=(cos(x-B),cosB),n=(cosx,-1/2）
f(x)=m●n=cos(x-B)cosx-1/2*cosB
=(cosxcosB+sinxsinB)cosx-1/2cosB
=cosBcos²x+sinBsinxcosx-1/2*cosB
=1/2cosB(1+cos2x)+1/2sinBsin2x-1/2*cosB
=1/2(cos2xcosB+sin2xsinB)
=1/2cos(2x-B)
∵f(π/3)=1/4
∴1/2cos(2π/3-B)=1/4
∴cos(2π/3-B)=1/2
∵0