题干应该是属于哪个取值范围吧!就按这个意思回答了.
答案：[3/4.3/2]
由题意知A B c =3B =π故B=π/3即A C =2π/3 现把A 角设为变量C角设为常量 有A￡(0 2π/3)sin?A sin?C=sin?A sin?(2π/3-A)=sin?A (√3/2cosa 1/2sinA)?=5/4sinA?3/4cos?A √3/2cosa*sinA=1-1/4cos2A √3/4sin2A=1 1/2sin(2A-π/6)而(2A-π/6)￡(-π/6,7π/6)故sin(2A-π/6)￡[-1/2.1]原式取值范围是[3/4.3/2]