1D Finite potential well: solutions with $\sinh$ and $\cosh$?
So I am studying the (one dimensional) quantum mechanical finite potential well defined by:
$$ V(x) = \cases{0, &|x|>a\cr -V_0, &|x|<a} $$
where $V_0>0$ is a real number.
I know that for $E>0$ there exists continuously many solutions, and for $-V_0<E<0$, there are even and odd solutions. If we denote by $I,II$ and $III$ the regions $x<-a,\;-a<xa$, respectively, then the even solutions are:
$$\psi_I(x)=A\exp(\kappa x)$$ $$\psi_{II}(x)=B\cos(kx)$$ $$\psi_{III}(x)=A\exp(\kappa x)$$
and the odd are:
$$\psi_I(x)=C\exp(\kappa x)$$ $$\psi_{II}(x)=D\sin(kx)$$ $$\psi_{III}(x)=-C\exp(\kappa x)$$
where $\kappa = \sqrt{\frac{2m|E|}{\hbar^2}}$ and $k=\sqrt{\frac{2m(V_0-|E|)}{\hbar^2}}$. The continuity conditions give the following conditions that must be satisfied by $\kappa$ and $k$ (and that gives the energy spectrum):
$$k\tan(ka)=\kappa\quad\quad\text{for the even solutions}$$ and $$k\cot(ka)=-\kappa\quad\quad\text{for the odd solutions}$$
Well, until now, everything's OK. But I was wondering about the case $E<-V_0$. In this case, the differential equation for region $II$ reads:
$$\frac{d^2\psi_{II}(x)}{dx^2}-\tilde{k}^2\psi_{II}(x)=0$$
where $\tilde{k}^2=\frac{2m(|E|-V_0)}{\hbar^2}$. 402
Well this equation has the following solutions (remember that the wave function must be even or odd):
$$\tilde{\psi_{II}}(x)=E\cosh(\tilde{k}x)\quad\quad\text{for the even case}$$
and
$$\tilde{\psi_{II}}(x)=F\sinh(\tilde{k}x)\quad\quad\text{for the odd case}$$
and the wavefunctions for the other regions don't modify. The analogous conditions I found for the energy spectrum are:
$$\tilde{k}\tanh(\tilde{k}a)=-\kappa\quad\quad\text{for the even case}$$
and
$$\tilde{k}\coth(\tilde{k}a)=-\kappa\quad\quad\text{for the odd case}$$
Then I asked myself: do solutions to the previous equations exist? I tried to verify this by making a graph. I set up $\frac{2m}{\hbar^2}=1,\;a=0.5$ and $V_0=1$. I verified that the equation in the even case is satisfied at $|E|\approx0.1893$, but this doesn't makes sense, since $V_0=1$ and $E<-V_0=-1$. The equation for the odd case is not satisfied.
The questions are: does all this stuff (the solution with $E<-V_0$) make any sense? Or all the bound state solutions (those with $E<0$) are given by the case $-V_0<E$? Or does the result in the previous paragraph prove that there aren't solutions with $E<-V_0$?
Sorry for the long question. 521