Graphical determination of energy eigenvalues (symmetrical potential well) It is about a particle with mass $m$ in a potential $V(x)$:

I want to do a graphical determination(at first only the symmetrical case) of the energy eigenvalues. I will show you my previous work: 
Since the independent Schrödinger Equation is $\varphi''+k^2(x)\varphi=0$ with $k^2(x)=\frac{2m}{\hbar^2 }\left[E-V(q)\right]$ there is following general solution for the wave-function 
$$\varphi(x) = \begin{cases} 0 &\mbox{if } |x| > b \\
\alpha_+e^{ikx}+\alpha_- e^{-ikx} & \mbox{if } -b < x < -a \\ \beta_+e^{\kappa x} + \beta_-e^{-\kappa x} & \mbox{if } -a < x < a \\ \gamma_+ e^{ikx}+\gamma_- e^{-ikx} & \mbox{if } a < x < b \end{cases}$$
(i.e. if $k^2 > 0 $ there is an oscillating pattern, which we have in zone I and III (classically allowed) and if $k^2 < 0$ an exponential behavior, which we only have in zone II(classically not-allowed))
We can simplify a few things for the symmetric solution: $\varphi(x) = \varphi(-x)$
So I only have to determine 
$$\frac{\varphi_3}{\varphi_3'}=\frac{\varphi_2}{\varphi_2'},         (1)$$
because this is the condition in this case, where the so-called Wronski Determinant is zero. In addition to we have some more conditions for combining the several functions in each Section I,II and II: $$\varphi_1(-b)=\varphi_3(b)=0$$ $$\varphi_1(-a)=\varphi_2(-a)$$ $$\varphi_2(a)=\varphi_3(a))$$
And of course the symmetrical charasteristic $\varphi_n(x) = \varphi(-x)$. To shorten this a here is the condition Eq. (1): $$\kappa \tanh(\kappa a) = k\cot[k(a-b)](2)$$

Let me compare this to another short example please: 
Imagine we have such potential: http://imgur.com/a/2PWWu"
And the Wronski Determinant would be zero here(symmetrical case), when $\kappa = k\tan(ka)$. By multiplying $a$ with this equation and by let being $\eta = \kappa\cdot a$ and $\xi = k\cdot a$ you will get $$\eta = \xi \tan \xi,$$ which brings you to following graphically solution of the energy eigenvalues of this potential well(ignore the dotted $cot$-function, it would be for the antisymmetrical case, the $tan$-function is for the symmetrical case):


So my question is now, how can I make my Eq. (2) only dependent from one value like in the example I showed above? Can someone give me a tip please, because I am clueless atm. 
 A: You potential is even so we expect on general grounds that solutions will divide into even and odds. I will discuss only the even case.
The candidate wavefunction is
\begin{align}
\psi(x)&=\left\{ {\renewcommand{\arraystretch}{1.25}\begin{array}{ll}
B\sin(k(b+x))&\hbox{if }-b\le x\le a\, ,\\
A\left(e^{\kappa x}+e^{-\kappa x}\right)&\hbox{if }-a< x < a\, ,\\
B\sin(k(b-x))&\hbox{if }a\le x\le b\, .\end{array}}\right.
\end{align}
It is certainly even under the transformation $x\to -x$, and it
satisfies the boundary condition that $\psi(\pm b)=0$ because the
wavefunction outside the well must be $0$.  This is just a generalization of your form where it is simpler to deal with boundary conditions, and where I've specialized to the even solutions.
This form assumes $E<V_0$. 
As always, define
$$
\kappa =\displaystyle{\sqrt{2m(V_0-E)\over \hbar^2}}\, ,\qquad
k=\displaystyle{\sqrt{2m E\over\hbar^2}}\, .
$$
The continuity of $\psi$ and $d\psi/dx$ at $x=a$ gives two
equations:
\begin{align}
A\left(e^{\kappa a}+e^{-\kappa a}\right)&=B\sin(k(b-a))\, ,\tag{1}\\
A\kappa\,\left(e^{\kappa a}-e^{-\kappa
a}\right)&=-B\,k\,\cos(k(b-a))\, .\tag{2}
\end{align}
The continuity of $\psi$ and its derivative at $x=-a$ gives the same
two equations.
Dividing (1) by (2) so as to eliminate $A$ and $B$, we obtain the
trancendental equation:
$$
\frac{1}{\kappa}\displaystyle\left({e^{\kappa a}+e^{-\kappa
a}\over e^{\kappa a}-e^{-\kappa
a}}\right)=-\frac{1}{k}\tan(k(b-a))\, .
$$
To bring this to a solvable form, we set
$$
\xi=\sqrt{2mV_0\,a^2\over \hbar^2}\, ,\qquad\qquad
z=\frac{E}{V_0}\, ,
$$
and (if I did my substitutions right)obtain the transcendental equation
$$
-\displaystyle{\frac{(e^{\xi\sqrt{1-z}}+e^{-\xi\sqrt{1-z}})}
{(e^{\xi\sqrt{1-z}}-e^{-\xi\sqrt{1-z}})\sqrt{1-z}}}=
\displaystyle{\frac{\tan(\xi\sqrt{z}(1-b/a))}{\sqrt{z}}}\, .
$$
Once you have obtained a numerical value for $\xi$ specific to you problem you can solve graphically for $z$ in the usual way.
The case of the odd solutions proceeds in the same general manner.
