Inverse circular function question!!

[madoscar65]

Hi guys,
I've been trying to find the domain and range of this equation of y=arccos(x^2). I've got the domain to be [-1,1], but I got the range to be [0,pi] but the answer says [0,pi/2]. Please help and please show me how you did it.

Thanks

[brightsky]

y = arccos(x^2) --> cos(y) = x^2

0 =< cos(y) =< 1 in this case so 0 =< y =< pi/2