Verified answer

This shouldn't be difficult:

r^2 = cosθ

r = cosθ

We can substitute cosθ directly if we wanted, or we could square the bottom equation and substitute r^2. Either way, we end up with:

(r, cosθ) = (0, 0) or (1, 1)

It doesn't make sense to consider θ where r = 0, so we only consider when θ = 1. Now it's simply a matter of solving:

cosθ = 1

And that, of course happens on even multiples of pi.

Pi/2 or 90deg