m=(√3sinx/4,1),n=(cosx/4,cos2x/4)
f(x)=m*n=√3sinx/4*cosx/4+cos2x/4
=（√3/2)*2sinx/4*cosx/4+(1/2)(1+cosx/2)
=cos(π/6)sinx/2+sin(π/6)cosx/2+1/2
=sin(x/2+π/6)+1/2
f(x)=1
sin(x/2+π/6)=1/2
cos(x+π/3)=cos[2(x/2+π/6)]=1-2(sin(x/2+π/6))^2=1-2*1/4=1/2
(2)
利用正弦定理：a=2RsinA,b=2RsinB,c=2RsinC,代入
（2a-c）cosB=bcosC 得到：
4RsinAcosB-2RsinCcosB=2RsinBcosC
消去2R,移项：
2sinAcosB=sinBcosC+cosBsinC=sin(B+C)=sinA
cosB=1/2,B=π/3 A+C=2π/3
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