 Solution

Writing $r=\sqrt{x^2+y^2}$, the absolute lower bound is $\cos r\ge -1$, attained exactly whenEvery point on these circles is a global and local minimum, but not a strict minimum: neighbouring points on the same circle have exactly the same value. At such a point the radial Hessian eigenvalue is one and the tangential eigenvalue is zero, illustrating why a degenerate second-derivative test does not exclude a minimum.

There are no other local minima: between successive multiples of $\pi$, moving radially changes the value monotonically. The positive even-radius circles are maxima. At the origin the apparently singular radial expression is smooth, since $\cos \sqrt{x^2+y^2}=1-(x^2+y^2)/2+O(r^4)$, and the origin is a strict local maximum.