What type of curve is described by $4(\cos{x}+\cos{y})-6(\cos{2x}+\cos{2y})+8\cos{x}\cos{y}=7$? Does the curve by the function $$4(\cos{x}+\cos{y})-6(\cos{2x}+\cos{2y})+8\cos{x}\cos{y}=7\\x=[-2\pi/3,2\pi/3],\; y=[-2\pi/3,2\pi/3]$$ belong to any known curve family? A collection of curves is found in Wikipedia (Link1,Link2). A large collection of $\approx 1000$ curves can be found here.

 A: Negative. I do not see the reason to have a name for the curve defined implicitly by the given condition
$$
4(\cos x + \cos y) 
- 
6(\cos 2x + \cos 2y )
+
8\cos x\cos y
=7\ ,
$$
where $x,y$ run only in the interval $I$ between the two points $\pm2\pi/3$.
However, we have a name for the curve isolated after using the substitutions
$$
\begin{aligned}
s &=\cos x\ ,\\
t &=\cos y\ ,
\end{aligned}
$$
which is the conic
$$
4(s+t) - 6(2s^2-1+2t^2-1) + 8st-7=0\ .
$$
Alternatively:
$$
12s^2 + 12t^2 - 8st - 4s - 4t -5 = 0\ .
$$
It is an ellipse, since the part in degree two $12s^2+12t^2-8st$ is always $\ge0$, and it is not degenerated
since its determinant is negative
$$
\begin{vmatrix}
12 & -4 &-2 \\
-4 & 12 &-2 \\
-2 & -2 & -5
\end{vmatrix}
=
2^2
\begin{vmatrix}
 6 & -2 &-1 \\
-2 &  6 &-1 \\
-2 & -2 & -5
\end{vmatrix}
=
2^4
\begin{vmatrix}
 3 & -1 &-1 \\
-1 &  3 &-1 \\
-1 & -1 & -5
\end{vmatrix}
=
2^4
\begin{vmatrix}
 4 & -4 & 0 \\
-1 &  3 &-1 \\
-1 & -1 & -5
\end{vmatrix}
=
2^6
\begin{vmatrix}
 1 & -1 & 0 \\
-1 &  3 &-1 \\
-1 & -1 & -5
\end{vmatrix}
=
2^6
\begin{vmatrix}
 1 &  0 & 0 \\
-1 &  2 &-1 \\
-1 & -2 & -5
\end{vmatrix}
=
2^7(-5-1)=-2^8\cdot 3<0\ .
$$
Since the switch $s\leftrightarrow t$ invariates the equation of the ellipse, one of its axes is $s=t$. This axis intersects the ellipse in the points with $s=t$ and $16s^2-8s-5=0$,
so $s=t=\frac 14(1\pm\sqrt6)$. The mid point between the two intersections is the center of the ellipse, $s=t=1/4$, and the other axis is the perpendicular in this point on the line $s=t$, so its equation is $s+t-\frac 12=0$.
The desmos picture looks like:

As seen from the picture, the lines $s=1$ and $t=1$ are tangent to the ellipse, indeed, setting $t=1$ in the equation leads to $0=12s^2-12s+30=3(2s+1)^2$, having a double solution. So the points $(s,t)=(1/2,1)$ and $(s,t)=(1,1/2)$ are the two tangent points.
This is relevant for us, since $\cos$ takes this maximal value $1$.
The corresponding $(x,y)$-points are $(x,y)=(\pm\pi/3, 0)$ and  $(x,y)=(0,\pm\pi/3)$, which are the four intersections of the graph with the coordinate axes. (Further points of the given graph can be traced back in this way to points of the ellipse.)

So the given curve has no name, but if a "name" should be given, it is the preimage of the above ellipse by the map $(s,t)\to(\cos s,\cos t)$.
