eqs=2ySinh[m] == cm(xSinh[m3] + yCosh[m3]);
Solve[eqs,x]
(* appears to solve quickly for x or y not both *)
x==Csch[m3] (c m y Cosh[m3] - 2 y Sinh[m])/(c m)
but with m1+n2 it does not solve for me and i have no time tonight for analyzing it.
but i observe in the above that i might have "solve x==cos(x)" or rather "solve x/y==cos(1/(x+y))" or moreso 0==y cos(1/(x+y))+x sin(1/(x+y)) situation which might be impossible except trivially, the solution may not be solvable for {x,y} (x or y), and the solution mathematica can give might be limited to expressing it in a different form (ie, with x~ ArcCos[Cos[x]]). i'd need to look further to know more.
if the above can't work for you, you can always plot and get a pseudo function for some given interval (sin cos curve) something like Interpolation[points]