# Solving the system $a+be^{-\lambda t_i}\cos(wt_i)=\epsilon_i$ (with $i=1,2$) for $\lambda$ and $w$

I have the following system of equations:

\begin{align} a+be^{-\lambda t_1}\cos(wt_1)&=\epsilon_1 \\ a+be^{-\lambda t_2}\cos(wt_2)&=\epsilon_2 \end{align}

$a$, $b$, $t_1$, $t_2$, $\epsilon_1$ and $\epsilon_2$ are given. I need to solve for $\lambda$ and $\omega$ and I know that $\lambda$ and $\omega$ are both real and positive.

How can I solve for those?

I did the following:

$$a+be^{-\lambda t_1}\cos(wt_1)=\epsilon_1\rightarrow\lambda=-\frac{1}{t_1}\cdot\ln\left(\frac{\epsilon_1-a}{b}\cdot\frac{1}{\cos(wt_1)}\right)$$

So:

$$a+b\exp\left(-\left\{-\frac{1}{t_1}\cdot\ln\left(\frac{\epsilon_1-a}{b}\cdot\frac{1}{\cos(wt_1)}\right)\right\}t_2\right)\cos(wt_2)=\epsilon_2$$

And only $\omega$ I do not know but I do not know how to solve the last equation.

• Have you given concrete values? Solving this equation explicitely is not possible. – Dr. Sonnhard Graubner Apr 6 '18 at 9:30
• @Dr.SonnhardGraubner The concrete values are measurement values. – Looper Apr 6 '18 at 9:30