解cos2A+3cosA+3cos(B-C)=1 =>3cosA+3cos(B-C)=1-cos2A =2sin^2A =3sin^2B+3sin^2C =>-3cos(B+C)+3cos(B-C)=3sin^2B+3sin^2C =>-(cosBcosC-sinBsinC)+(cosBcosC+sinBsinC)=sin^2B+sin^2C =>sin^2B+sin^2C-2sinBsinC=...