在三角形ABC中,已知2sin^2A=3sin^2B+3sin^2C,cos2A+3cosA=3cos(b-c)=1,求：a:b:c
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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=0
=>(sinB-sinC)^2=0
=>sinB=sinC
∵B+Csin^2A=3sin^2B
=>sinA=√3*sinB
∴a:b:c=√3:1:1
=>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=0
=>(sinB-sinC)^2=0
=>sinB=sinC
∵B+Csin^2A=3sin^2B
=>sinA=√3*sinB
∴a:b:c=√3:1:1
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