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)=1/2cos2x−2/3cosx+1/2,求f(A)的取值范围.

问题描述:

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)=

1
2
cos2x−
2
3
cosx+
1
2
,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≠...