Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse? The solution set of $\cos(x) + \cos(y) - \cos(x + y) = 0$ looks like an ellipse. Is it actually an ellipse, and if so, is there a way of writing down its equation (without any trig functions)?
What motivates this is the following example. The solution set of $\cos(x) - \cos(3x + 2y) = 0$ looks like two straight lines, and indeed we can determine the equations of those lines.
$$
\begin{align}
\cos(x) &= \cos(3x + 2y) \\
\implies x &= \pm (3x + 2y) \\
\implies x + y &= 0 \text{ or } 2x + y = 0
\end{align}
$$
Can we do a similar thing for the first equation?

we have $$\begin{align}  0 &= \cos x +\cos y -\cos(x + y)\\
&=\cos x +\cos y -\cos x \cos y+ \sin x \sin y\\
&=(1-\cos x)\cos y+\sin x \sin y+\cos x\\
&=2\sin^2 (x/2)\cos y+2\sin(x/2)\cos(x/2)\sin y + \cos x\\
&=2\sin(x/2)\left(\sin (x/2)\cos y+\cos(x/2)\sin y)\right)+\cos x\\
&=2\sin(x/2)\sin (x/2 + y)+\cos x\\
\end{align} $$
therefore $$y=-\frac x2-\sin^{-1}\left(\frac{\cos x}{2\sin (x/2)}\right)+2k\pi,-\frac x2 +\sin^{-1} \left(\frac{\cos x}{2\sin (x/2)}\right)+2k\pi + \pi   .\tag 1 $$
when i graphed $(1),$ what i see kind of ellipses all running parallel to the line $y = -x/2.$
