设A-B=x,B-C=y,C-A=z,则x+y+z=0
2cosx+2cosy+2cosz
=2cos[(x+y)/2+(x-y)/2]+2cos[(x+y)/2-(x-y)/2]+2cos(x+y)
=4cos^2[(x+y)/2]+4cos[(x-y)/2]*cos[(x+y)/2]-2
=-3
4cos^2[(x+y)/2]+4cos[(x-y)/2]*cos[(x+y)/2]+1=0
由判别式得：cos[(x-y)/2]>=1
所以cos[(x-y)/2]=1
4cos^2[(x+y)/2]+4cos[(x+y)/2]+1=0
cos[(x+y)/2]=-1/2
cos（x+y）=-1/2
x+y=A-C
cos（A-C）=-1/2
cos（2A-2C）=-1/2