Given function,
f : R → R, f(x) = cos (x + 2)
Putting x = 2π
f(2π) = cos (2π + 2) = cos (2)
Putting x = 0
f(0) = cos (0 + 2) = cos 2
Since, only one image is obtained for 0 and 2π, so function is not one-one. Hence, function is not one-one onto.
Hence, f-1 : R → R does not exist.