在三角形ABC中，已知2sin^2A=3sin^2B+3sin^2C,cos2A+3cosA=3cos(b-c)=1,求：a:b:c

1-2sin^2A+3cosA+3cos(B-C)=1
2sin^2A=3cosA+3cos(B+C)
3sin^2B+3sin^2C=3cosA+3cosBcosC+3sinBsinC
3(sinB-sinC)^2=3cosA+3cos(B+C)
∴（sinB-sinC)^2=0
∴sinB=sinC
∴sinA=根号3sinB=根号3sinC
∴sinA:sinB:sinC=a:b:c=根号3：1：1

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+C<180
∴B=C 2sin^2A=3sin^2B+3sin^2C=6sin^2B
=>sin^2A=3sin^2B
=>sinA=√3*sinB
∴a:b:c=√3:1:1