在△ABC中,角A,B,C所对的边分别为a,b,c,且asinB-bcosC=ccosB.
(Ⅰ)判断△ABC的形状;
(Ⅱ)若
f(x)=cos2x−cosx+,求f(A)的取值范围.
(Ⅰ)法1:∵asinB-bcosC=ccosB,由正弦定理可得:sinAsinB-sinBcosC=sinCcosB.即sinAsinB=sinCcosB+cosCsinB,∴sin(C+B)=sinAsinB,∵在△ABC中,A+B+C=π,即C+B=π-A,∴sin(C+B)=sinA=sinAsinB,又sinA≠...