What type of curve is described by $\cos{x}+\cos{x}\cos{y}+\cos{y}=0$? Does the curve by the function $$\cos{x}+\cos{x}\cos{y}+\cos{y}=0\\x=[-2\pi/3,2\pi/3],\; y=[-2\pi/3,2\pi/3]$$ belong to any known curve family? Examples of curves can be found in Wikipedia (Link1, Link2) but the longest list with $\approx 1000$ curves you find here.

 A: This implicit equation can be written
$$(1+\cos x)(1+\cos y)=1$$
Using a classical trigonometric formula:
$$4(\cos(x/2)\cos(y/2))^2=1$$
$$2\cos(x/2)\cos(y/2)=\pm 1$$
Due to invariance with respect to changes $x \to -x, y \to -y$ (in connection with symmetries with respect to coordinate axes), one can reduce the study to the first quadrant with equation:
$$\cos(y/2)=\frac{1}{2 \cos(x/2)}$$
finally giving a cartesian equation for the curve in the first quadrant:
$$y=2 \arccos \left(\frac{1}{2 \cos(x/2)}\right) $$
This form doesn't evoke more anything known than the implicit form (but see the Edit below).
A first check : if $x=0$ we get $y=2 \frac{\pi}{3}$ which is the point $(0,2 \frac{\pi}{3}\approx 2.09)$ on your curve.
Remark: there is also another symmetry with respect to line bissector $y=x$.
Important edit: With the help of Geogebra, I have found a very good fit of this curve with the following so-called "squircle" with equation
$$|x|^{2.42}+|y|^{2.42}=5.95\tag{1}$$
as we can see on the following representation (the red curve with equation (1) hides almost completely the initial curve, in black).
It would be interesting to understand why such a good fit exists.

A: You may say $2$-D version or cross-section of Fermi-surfaces (especially for cubic system).
See the posts here, here and also here.


Addendum
See also an article in Mathematical Intelligencer here.
A: It approximates to  fourth, sixth, eighth order ellipses depending on number of chosen terms in binomial series.
Taking the lowest second order terms only
$$ (1+\cos x)(1+\cos y)=1 ; \to x^2+y^2= 2$$
is a circle radius $\sqrt 2 $ centered at origin.
If instead of $\sqrt 2 $ we take on higher values for the exponent, i.e., as $ 2n \to \infty$... so that
$$x^{2000}+y^{2000}\to (\sqrt 2)^{2000}$$
approximating to the unit square forming more pointed vertices at corners of the figures in the process.
Iirc, a reasonably large value of $2n$ is chosen to approximate aircraft window cutout contours in aircraft fuselage usage adapted to out-of-plane 3d profiles.
