Is it possible to calculate the negative area of $f.g.h$ analytically? I have the three two-variable curves $f,g,h$ with $0\leq x\leq \dfrac {\pi}{2} $ and $0 \leq y \leq \pi:$
\begin{align*}
f&=2 \cos (x-2 y)+\cos (x)\\
g&=7\cos (x-y)+ \cos (x+2y)\\
h&=3\cos (x-y)+ \cos (x).
\end{align*}
I want to calculate analytically the area in the $xy$ plain in which $f.g.h$ is negative.
My question:

Is there any hope to calculate this area analytically?

I really appreciate any hints and comments.
 A: This is the best way that I have found
Can we paramtersize the curves $f(x,y) = 0, g(x,y) = 0, h(x,y) = 0$?
If we can, then we can use contour integrals to find the area of the blue regions.
$f(x,y) = 0$
let:
$u + v = x-2y\\
u - v = x$
$f(u,v) = 2\cos (u+v) + \cos (u-v) = 0\\
2\cos u\cos v - 2\sin u\sin v + \cos u\cos v + \sin u\sin v = 0\\
3\cos u\cos v = \sin u\sin v\\
\tan u\tan v = 3\\
u = \arctan (3\cot v)\\
x = \arctan (3\cot v) - v, y = -v$
To get rid of the negative signs, we can swap the sign on $v$ and carry it though the arctan function.
$(x,y) = (-\arctan (3\cot v) + v, v)$
Plotting this curve over the curve of $f(x,y)$ we see that
$(x,y) = \left(-\arctan (3\cot v) + v , v\right)$ with $v \in (\frac {\pi}{3}, \frac {2\pi}{3})$ corresponds to the lower of the two curves, and  $(x,y) = \left(-\arctan (3\cot v) + v - \pi, v\right)$ with the higher curve.
The lets us find the area of the upper region is:
$A = \int_{\frac {2\pi}{3}}^{\pi} y \frac {dx}{dv} dv  + \int_0^{\frac {\pi}{2}} \pi \ dx$
$g$ looks a little easier to work with.
We can use a similar technique to parameterize $h$
And I notice that the intersection of $f$ and $g$ is on the line $y=2x$  Not sure if that will be useful or not.
I hope this gets you started.
