= Solution
Writing $r=\sqrt{x^2+y^2}$, the absolute lower bound is $\cos r\ge-1$, attained exactly when
$$
\boxed{x^2+y^2=[(2n+1)\pi]^2,\qquad n=0,1,2,\ldots.}
$$
Every 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.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-radial-minima-and-saddle-contours.png]
{title=Nonisolated radial minima and a two-lobed saddle contour}
{description=Original level plots. The left panel marks whole circles of global minima; the right marks the saddle at the origin, two global maxima and the exact zero contour.}
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