First translate the relation so that one of the ovals is centered on the origin.

x->x+pi, y->y+pi 

This changes the relation into cos(x+y) = -cos(x) - cos(y).

Then rotate the relation 45 degrees.

x-> (x-y)/sqrt(2), x-> (x+y)/sqrt(2)

This changes the relation into cos(2x/sqrt(2)) = -cos((x-y)/sqrt(2)) - cos((x+y)/sqrt(2))

Then scale the figure by a factor of sqrt(2) on both axes.

x->sqrt(2)*x, y->sqrt(2)*y

This changes the relation into cos(2x) = -cos(x-y) - cos(x+y)

Note this last transformation will have half the area of the original.

Apply basic trig identities to cos(2x) = -cos(x-y) - cos(x+y) eventually simplifies into cos(y) = 0.5*sec(x) - cos(x)

At y=0 then x=arccos((-1+sqrt(3))/2) and at x=0 then y=2*pi/3

Then focus on the first quadrant, which is one fourth of the figure.  Then y = f(x) = arccos(0.5*sec(x) - cos(x))

Then the area of the original figure can be evaluated by calculating 8 times the integral of f(x) from 0 to arccos((-1+sqrt(3))/2).

I could not figure out an antiderivative, so a numerical integration of this gave me a final answer of 15.73736.  This is quite close to the answer Charlie derived by making assumption that the figures were ellipses.

I do want to see what a Monte-Carlo simulation does over the original function inside the square (0,0) to (2pi,2pi).