sin^22B+sin2BsinB+cos2B=1
化简2(sinB)^2*[2(cosB)^2+cosB-1]=0
因为sinB>0 则(2cosB-1)(cosB+1)=0
因cosB>-1
故cosB=1/2 B=60°
则A+C=A=180°-B=120° (A+C)/2=60°
A-C=2A-120° (A-C)/2=A-60°
由正弦定理a/sinA=c/sinC=b/sinB=3/sin60°=2√3
则a+c=2√3(sinA+sinC)=√3sin(A+C)/2cos(A-C)/2
=√3sin60°cos(A-60°)=(3/2)cos(A-60°)
可见当且仅当A=60°时,a+c最大=3/2