solve system of two trigonometric equations I have two equations:
$$f(x) = x^2/k^2$$
and 
$$ g(x) = a \cdot sinh(bx)$$
The derivatives are therefore
$$ f'(x) = 2x/k^2$$ and $$g'(x) = ab \cdot cosh(bx)$$
Now for given values of $k$ and $x$ (for example: $k=0.42$ and $x=7$) I want to find values of $a$ and $b$ such that the values and also the slopes of $f$ and $g$ are identical,
which should be doable by solving the set of two equations $f=g$ and $f'=g'$ simultaneously.
From equation $f$ I can get
$$ a= \frac{x^2}{k^2 \cdot sinh(bx)}$$
and insert that in $g$,
or from $g$ I can get
$$ a= \frac{2x}{bk^2 \cdot cosh(bx)}$$
and insert that in equation $f$,
but then I am still not able to solve the resulting single equation.

Since you have $a$ in both equation, you can eliminate it and the equation becomes
$$
\frac{x}{\sinh(bx)}=\frac{2}{b\cosh(bx)}
$$
that is,
$$
bx=2\tanh(bx)
$$
Consider the function
$$
f(t)=2\tanh t-t
$$
that we can study for $t\ge0$. The derivative is
$$
f'(t)=\frac{2}{\cosh^2t}-1=\frac{2-\cosh^2t}{\cosh^2t}
$$
which vanishes at $\cosh t=\sqrt{2}$ and is positive in the interval $[0,\operatorname{arcosh}\sqrt{2}]$. Moreover, $\lim_{t\to\infty}f(t)=-\infty$. Thus the equation $f(t)=0$ has a single positive solution $t_0$, that can be determined numerically; an approximate value is $1.9189$. 
Once you have found it, you get
$$
b=\frac{t_0}{x}
$$
and you can compute the value of $a$.
