答：
-π/2<=x<=0时,0<=cosx<=1,0<=-x<=π/2
y=arccos(cosx)=arccos[cos(-x)]=-x
0<=x<=π/2时：
y=arccos(cosx)=x
所以：
-π/2<=x<=0时,y=arccos(cosx)=-x
0<=x<=π/2时,y=arccos(cosx)=x