What type of curve is described by $\cos{x}+\cos{x}\cos{y}+\cos{y}=0$?

Solution 1:

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.