explicit solution for transcendental equation Does anyone knows whether there is an explicit, analytical solution for transcendental equations of the form $A x + B \tanh(C x) + \coth(x) = 0$, where $A, B$, and $C$ are positive real constants?
 A: The Burniston-Siewert approach for anaytical solution of similar transcendental equation is by integral representations.
In :
http://www4.ncsu.edu/~ces/publist.html
You can find a lot of transcendental equations solved by such integrals, therefore you could adapt this method to your custom problem.
?? HINT ??
For a suitable complex path $\gamma$, x(A,B,C) can be written as:
$$x(A,b,C)=\frac{1}{2\pi i} \oint_\gamma z\frac{A+\frac{BC}{\cosh(Cz)^2}-\frac{1}{\sinh(z)^2}}{Az+B\tanh(Cz)+\coth(z)}dz$$
Historical Note
In :
"The $W_t$ Transcendental Function and Quantum Mechanical Applications", V. E. Markushin, R. Rosenfelder, A. W. Schreiber, http://arxiv.org/abs/math-ph/0104019v2
the equation $$x \tan(x) =y$$ has been investigated and the Lambert-like function $W_t(x)$ has been proposed. Such function is linked to the so called Lambert r-$W$ (i.e.  extension of Lambert series based on Laguerre polynomials).
A: This is a comment on GEdgar's ($C=0$) version and the equation
$$
Ax+\coth(x)=0,
$$
(i.e. Find the points of intersection of the curve $y=\coth(x)$
and straight lines through the origin. 
Existence of real solutions requires $A$ to be of an appropriate sign.)
This can be solved using the functions in
I. Mező and A. Baricz,
On the generalization of the Lambert $W$ function with applications in theoretical physics.
arXiv:1408.3999v1.pdf (18 Aug 2014).
The starting point to using their functions is to rewrite the equation as
$$
\exp(2x)= \frac{x-1/A}{x+1/A}
$$
and then with $X=2x$ define
$$
f(X)= \frac{X+2/A}{X-2/A}\exp(X).
$$
The inverse function for f is defined in the arXiv paper (and earlier papers referenced there), and hence  the solution(s) for $f(X)=1$.
Remark. The inverse Langevin function can also be written in terms of the functions in their arXiv paper.
A: This is a comment
in response to giorgiomugnaini's
comment about an exact solution.
I an making it an answer
since entering the MathJax is much easier.
The "solution" 
to the equation
$ 0 = a + bx-x \tanh^{-1}(\frac1{x})$
is given, 
in equation (16),
when $a < 0$, by
$\frac1{x}
=2+\frac{1-a}{b}-\frac1{\pi}
\int_{-1}^1 \tan^{-1}\left( \frac{\pi t}{2(a+bt-t\tanh^{-1}(t))}\right) dt
$.
In view of the
very non-elementary nature
of that integral,
this does not seem to me
to be much of a help.
A: Here is a Lagrange Reversion solution for the larger root of a more general equation:
$$x-(a \tanh( b y)+c \coth(y))=y\implies y=x+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\frac{d^{n-1}(a\tanh(b x)+c\coth(x))^n}{dx^{n-1}}\mathop=^{r=\frac ac} x+\sum_{n=1}^\infty\frac{(-c)^n}{n!}\frac{d^{n-1}(r\tanh(b x)+\coth(x))^n}{dx^{n-1}} $$
Compare results here and here. Now to use the binomial theorem:
$$\frac{d^{n-1}(r\tanh(b x)+\coth(x))^n}{dx^{n-1}} =\sum_{k=0}^n \binom nk r^k\frac{d^{n-1}}{dx^{n-1}}\coth^{n-k}(x)\tanh^k(b x)$$
and somehow take the $n$th derivative of the hyperbolic functions using the single exponential definiton. More will be added
