在△ABC中,角A,B,C的对边分别为a,b,c,且bcosC=3acosB-ccosB.
(Ⅰ)求cosB的值;
(Ⅱ)若
•=2,且
b=2,求a和c的值.
(I)由正弦定理得a=2RsinA,b=2RsinB,c=2RsinC,则2RsinBcosC=6RsinAcosB-2RsinCcosB,故sinBcosC=3sinAcosB-sinCcosB,可得sinBcosC+sinCcosB=3sinAcosB,即sin(B+C)=3sinAcosB,可得sinA=3sinAcosB.又sinA≠0...