Seeking an explicit solution (using Lambert W?) for $\lambda$ in $2\lambda = \sinh(\lambda)+(1+2\beta)(\cosh(\lambda)-1)$ Let $\beta\in \mathbb{R}$, I'm looking for an explicit solution in terms of $\lambda$ to the following equation:
$$2\lambda = \sinh(\lambda)+(1+2\beta)(\cosh(\lambda)-1)$$
It seems, numerically, that for $\beta\in [-1,0]$ the equation has three real solutions $(0,\lambda^*_1,\lambda^*_2)$ and for $\beta>0$ or $\beta<-1$ there are two solutions $(0,\lambda^*_3).$
Could there exist expressions of these solutions in terms of Lambert functions by any chance? See my answer below for a conjecture about the general form.
 A: $$(\beta+1)e^\lambda+\beta e^{-\lambda}-2\lambda-2\beta-1=0$$
The equation is not in a standard form: $e^{-\lambda}$ disturbs. We multiply the equation by $e^\lambda$ therefore.
$$(\beta+1)\left(e^\lambda\right)^2-2\lambda e^{\lambda}-2\beta e^{\lambda}-e^{\lambda}+\beta=0$$
The standard form $c_1\left(e^{\lambda}\right)^2+c_2e^{\lambda}+c_3\lambda e^{\lambda}+c_4\lambda+c_5=0$ has explicit solutions in terms of elementary functions and/or LambertW e.g. for $c_1=c_3=0$, $c_1=c_4=0$, $c_2=c_3=0$, $c_3=c_4=0$, $c_4=c_5=0$ and $c_1=c_2=c_5=0$.
For $\beta=-1$, we get the solutions $\lambda=0$ and $λ=W\left(−1,−\frac{1}{2}e^{-\frac{1}{2}}\right)+\frac{1}{2}$.
For $\beta=-0.5$, we get i.a. the solution $\lambda=0$.
For $\beta=0$, we get the solutions $\lambda=0$ and $λ=−W\left(−1,−\frac{1}{2}e^{-\frac{1}{2}}\right)−\frac{1}{2}$.
For $\beta=0.5$, we get i.a. the solution $\lambda=0$.  
A: Too long for a comment.
I really wonder how you did arrive to this conjecture (I guess that crazy is appropriate) but it seems to be numerically correct only for the range $-1.5 \leq \beta \leq 0.5$ if using the $W_0(.)$ branch.
From a numerical point of view, excluding the trivial $\lambda=0$, I should rewrite the equation as
$$\beta =\frac{2 \lambda -\sinh (\lambda )-\cosh (\lambda )+1}{2 (\cosh (\lambda )-1)}$$ and the rhs looks like an hyperbola with horizontal asymptotes $0^-$ when $\lambda \to - \infty$ and $-1^+$ when $\lambda \to + \infty$. This explains one root for $\beta >0$ and two roots when $-1< \beta <0$.
Concerning approximations, using Taylor expansions around $\lambda=0$, we have (this seems to be  rather good )
$$\beta=\frac{1}{\lambda }-\frac{1}{2}-\frac{\lambda }{4}+\frac{7 \lambda
   ^3}{720}+O\left(\lambda ^4\right) $$
A: Too long for a comment
Hi Claude,
Here is how I found this conjecture. Let me tell you first that the following calculations do not seem to make sense, so do not consider it at a rigorous level. 
We recall that want to solve the following equation for any $\alpha \in \mathbb{R}$ (this is an equivalent form as above):
\begin{equation}
    (1+\beta) (e^\lambda -1)-\lambda=\beta (1-e^{-\lambda})+\lambda
\end{equation}
We write it as
\begin{equation}
    \frac{1+\beta}{1+2\beta} e^\lambda +\frac{\beta}{1+2\beta}e^{-\lambda}-(\frac{\lambda}{\beta+\frac{1}{2}}+1)=0
\end{equation}
We introduce a Bernoulli random variable $p$ which takes value $+1$ with probability $  \frac{1+\beta}{1+2\beta}$ and $-1$ with probability $  \frac{\beta}{1+2\beta}$.
Note that this might imply some conditions on $\beta$ for the random variable to be properly defined.
We rewrite the equation to solve as 
\begin{equation}
\mathbb{E}_p\left[e^{p \lambda}-(1+\frac{\lambda}{\beta+\frac{1}{2}}) \right]=0 
\end{equation}
We now solve the reduced equation 
\begin{equation}
    e^{p \lambda}-(1+\frac{\lambda}{\beta+\frac{1}{2}}) =0
\end{equation}
The solution is 
\begin{equation}
    \lambda_p=-(\beta+\frac{1}{2}) -\frac{W\left(- p( \beta +\frac{1}{2})  e^{-p\left(\beta +\frac{1}{2}\right)}\right)}{p}
\end{equation}
There might be multiple solutions due to the two real branches of the Lambert $W$ function.
My conjecture is then that the solution to my first problem is the expected value of $\lambda_p$
\begin{equation}
    \lambda=\mathbb{E}_p[\lambda_p]=-(\beta+\frac{1}{2}) +\frac{\beta  W\left( e^{\beta +\frac{1}{2}} ( \beta
   +\frac{1}{2})\right)- (\beta +1) W\left(- e^{-\beta -\frac{1}{2}} (
   \beta +\frac{1}{2})\right)}{2 \beta +1}
\end{equation}
There is no mathematical ground for this conjecture but somehow it seems to do the job. The question is why ?
A: The equation can easily written in the form
$$
(\beta+1)e^{2\lambda}-2\lambda e^{\lambda}-2\beta e^{\lambda}-e^{\lambda}=-\beta.\tag 1
$$
Assume the function $\lambda(x)$ is such
$$
\lambda e^{\lambda}=P(\beta),
$$
where $P(x)$ will be set later. Then equation (1) becomes
$$
(\beta+1)e^{2\lambda}-(2\beta+1)e^{\lambda}=2P(\beta)-\beta.\tag 2
$$
Hence solving (2) with repect to $e^{\lambda}$ we get
$$
e^{\lambda}=Q(\beta):=\frac{2\beta+1\pm\sqrt{1+8(\beta+1)P(\beta)}}{2(\beta+1)}.\tag 3
$$
Multyplying both sides of (3) with $\lambda$, we get
$$
P(\beta)=\frac{2\beta+1\pm\sqrt{1+8(\beta+1)P(\beta)}}{2(\beta+1)}\lambda.
$$
Hence
$$
\lambda=W\left(P(\beta)\right)=\frac{2(\beta+1)P(\beta)}{2(\beta+1)\pm\sqrt{1+8(\beta+1)P(\beta)}-1}
$$
Inverting $P(\beta)$ and solving with repect to $P^{(-1)}(\beta)$, we get :
$$
P^{(-1)}(\beta)=-\frac{\beta(\beta-W(\beta)-2W(\beta)^2)}{(\beta-W(\beta))^2},\tag 4
$$
where $W(z)$ is Lambert function i.e. $W(z)e^{W(z)}=z$. Hence if $L(\beta)$ is ''the known function''
$$
L(\beta):=-\frac{\beta(\beta-W(\beta)-2W(\beta)^2)}{(\beta-W(\beta))^2},\tag 5
$$
its inverse will be $P(\beta)$. Hence
$$
P(\beta)=L^{(-1)}(\beta)
$$
and the solution of (1) is
$$
\lambda=W\left(P(\beta)\right)=W\left(L^{(-1)}(\beta)\right).\tag 6
$$
NOTES. The solution you have found
$$
\lambda=-\left(\beta+\frac{1}{2}\right)+\frac{\beta}{2\beta+1}W\left(e^{\beta+\frac{1}{2}}\left(\beta+\frac{1}{2}\right)\right)-\frac{\beta+1}{2\beta+1}W\left(-e^{-\beta-\frac{1}{2}}\left(\beta+\frac{1}{2}\right)\right)\tag 7
$$
by using the identity
$$
W(xe^{x})=x\tag 8
$$
becomes
$$
\lambda=-\left(\beta+\frac{1}{2}\right)+\frac{\beta}{2\beta+1}\left(\beta+\frac{1}{2}\right)-\frac{\beta+1}{2\beta+1}\left(-\beta-\frac{1}{2}\right)=0
$$
and is valid for all $\beta$. 
CONTINUING
I have ploted the solution you conjectured. (In Mathematica Program), $x\in\textbf{C}$, $k\in\textbf{Z}$.
$$
W[k,x]:=ProductLog[k,x]
$$
$$
S[k,x]:=-(x+1/2)+x/(2x+1)W[k,(x+1/2)Exp[x+1/2]]-(x+1)/(2x+1)W[k,(-x-1/2)Exp[-x-1/2]]
$$
Ecxept for the value $k=0$, where it becomes $\lambda=S[0,x]=0$ (trivialy set $\lambda=0$ in (1)), for all $x\in [-3/2,1/2]$; the formula $\lambda=S[k,x]$ is not a solution. Hence there is no conjecture and all this by numerical verification.
