A.A=B B.A=C C.B=C D.A=B=C
C
解析:由sinB·sinC=cos2=1+,
得2sinBsinC=1+cosA,
化简cos(B-C)-cos(B+C)=1+cosA,
又cos(B+C)=-cosA,
∴cos(B-C)=1,
∴B-C=0,即B=C,选C.