The question reads Let f(x) = 6 cos^-1 x + 2009pi
Find x values such that f^-1 (x) = (sq rt 3)/2
So I am thinking that with the cos ^-1 that the range for x is 0 to pi so that it is a one to one function.
Then what I am thinking is to find
6 cos^-1 x + 2009pi = (sq rt 3)/2 to find out what x leads to (sq rt 3)/2
But where do I go from here? I don't know how to handle the cos ^-1 x.