△ABC中 ，a*cosC+(根号3)a*sinC-b-c=0。1.求A 2.若a=2，S△ABC=根号3，求b,c

解：(1)∵a*cosC+√3a*sinC-b-c=0
∴a*cosC+√3a*sinC=b+c
由正弦定理a/sinA=c/sinC
与余弦定理cosC=(a²+b²-c²)/2ab，得
(a²+b²-c²)/2b+√3c*sinA=b+c
则 a²+b²-c²+2√3bc*sinA=2b²+2bc
∴2√3bc*sinA-2bc=b²+c²-a²
即 2√3bc*sinA-2bc=2bc*cosA
∴√3sinA-cosA=1
∴sin(A-π/6)=1/2
∴A-π/6=π/6或A-π/6=5π/6
∴A=π/3或A=π(舍去)
(2)S=(1/2)bc*sinA
故 bc=2S/sinA
=2×√3÷sin(π/3)
=4 ①
又cosA=(b²+c²-a²)/2bc
∴b²+c²=a²+2bc*cosA
=2²+2×4×cos(π/3)
=8 ②
由①②，得b=c=2