i think we may be over complicating it a bit, so rather than equating fx and fy, how about noticing the following
fx=cos(x+y)+cos(x)=0
gives
cos(x)=-cos(x+y)
similarly for fy
fy=cos(x+y)+cos(y)=0
gives
cos(y)=-cos(x+y)
so it is true that for the above to be satisfied requires cos(x)=cos(y), though that only means fx=fy, not fx=fy=0, which is the overall goal.
however it will be instructive to first consider cos(x)=cos(y), to solve this, consider a graph of cos(x), clearly it repeats on every 2.pi, so if y is s solution, so is y+2.pi
Now confine x to (-pi,pi), there will be 2 solutions to cos(x)=cos(y), clearly y=x and using the fact cos is symmetric cos(y)=cos(-y), gives y=-x.
Thus all the solutions are of the form:
y = x+2.pi
y = -x+2.pi
Now going back to the main requirement for fx=fy=0:
cos(x)=-cos(x+y)
cos(y)=-cos(x+y)
can you do a similar process to find allowable (x,y) pairs?