在三角形abc中,acosc+根号3asinc-b-c=0 求角a 求a=2时 b+c的取值范围

acosc+√3asinc-b-c=0
sinAcosC+√3sincsinA-sinB-sinC=0
sinAcosC+√3sincsinA=sinB+sinC=sin（A+C）+sinC=sinAcosC+sinCcosA+sinC
sinC≠0
√3sinA-cosA=1
sin(A-30°)=1/2
A-30°=30
A=60°,B+C=120
2)b+c=acosc+√3asinc=a(cosc+√3sinc)=2asin(c+π/6),a=2
b+c=4sin(c+π/6),
C∈(0,2π/3)
c+π/6∈(π/6,5π/6)
1/2b+c∈(2，4]

acosC+√3asinB-b-c=0
利用正弦定理 a/sinA=b/sinB=c/sinC
sinAcosC+√3sinAsinC-sinB-sinC=0
∵ sinB=sin(A+C),
sinAcosC+√3sinAsinC-sin(A+C)-sinC=0
sinAcosC+√3sinAsinC-sinAcosC-cosAsinC-sinC=0
√3sinAsinC=sinC+cosAsinC
√3sinA=1+cosA
2√3sin(A/2)cos(A/2)=2cos²(A/2)
√3tan(A/2)=1
tan(A/2)=√3/3
∵0