(New answer)
Some of this overlaps with the answer by Jayesh Badwaik, but it does not hurt to see two people's perspective.
Concerning the sequence $t_k$, let's begin with a simple observation based on Rolle's theorem: there exists $t$ strictly between $t_k$ and $t_{k+1}$ where the derivative of $e^t(g\cos\omega t\pm b)$ vanishes. This means $g\cos\omega t-g\omega \sin\omega t\pm b=0$. If $g\sqrt{1+\omega^2}\le b$, the derivative never changes its sign; therefore in this case you don't even have $t_1$. If $g\sqrt{1+\omega^2}> b$, there are infinitely many solutions which form a periodic pattern. This does not tell us that the sequence $t_k$ is infinite, but if it is infinite, then it grows at least linearly: $t_k\ge Ak+B$.
When $g>b$, the function $e^t(g\cos\omega t\pm b)$ is oscillating with increasing amplitude, roughly between $\pm g e^t$. It is not hard to see that in this case there is always $t_{k+1} \in (t_k, t_k+2\pi/\omega]$. So, the sequence $t_k$ is infinite and it grows at linear rate. Notice that this contradicts your observation that the sequence stops with $b=0.8$ and $g=1.75$. I suspect that you ran into overflow issues due to the exponential growth of the function.
The case $g=b$ is about the same as $g>b$: here the function does not change sign, but it returns to zero infinitely often. The sequence $t_k$ is infinite and grows at linear rate.
In the interval $g< b< g\sqrt{1+\omega^2}$ the elementary observations above do not suffice.
Let's step back from the case-by-case analysis and try to see more structure. Let $F(t)=e^t(g\cos\omega t+ b)$. The equation $F(t_{k+1})=F(t_k)$ is invariant under adding an integer multiple of $2\pi/\omega$ to both $t_k,t_{k+1}$. So, we can choose the starting point $t_0$ to be in an interval of length $2\pi/\omega$ of our choice: it appears to be convenient to take the interval between two consecutive local minima of $F$. We can also choose to subtract a multiple of $2\pi/\omega$ from subsequent points $t_k$ to avoid the exponential growth and associated numerical issues.
On every other step you want to solve a modified equation with $-b$ in place of $b$: namely, $e^{t_{k+1}} (g\cos(\omega t_{k+1}) - b) = e^{t_k} (g\cos(\omega t_k) - b)$. But this equation can be put in terms of the same function $F$ as follows: $F(t_{k+1}-\pi/\omega)=F(t_k-\pi/\omega)$. I could as well use $+\pi/\omega$ since the integer multiples of $2\pi/\omega$ do not matter. But it seems advisable to jump to the left rather than to the right, to keep the sequence from growing. Here is the algorithm:
- Start with $s_0$.
- Find the smallest $s_{k+1}>s_{k}$ such that $F(s_{k+1})=F(s_k)$. If there is no solution, stop.
- Replace $s_{k+1}$ with $s_{k+1}-\pi/\omega$.
- Increment $k$ and return to step 2.
This process continues for as long as the original process with $t_k$, and $s_k$ agrees with $t_k$ up to an integer multiple of $\pi/\omega$.
This algorithm can be useful not just for computations, but also for understanding of what can happen. For example, if $F(s)=F(s+\pi/\omega)$, and this common value of $F$ does not occur in between, then starting with $s_0=s$ we get $s_1=s_2=\dots=s$, and the process goes on forever. Here is a concrete example: $f=e^t(\cos 3t+2)$; the starting point is $(1/3)\cos^{-1}(2(e^{\pi/3}-1)/(e^{\pi/3}+1))$. This example shows that infinite process is possible even if $g<b$.
You can get more information by following the algorithm yourself with a printout of the graph of $F$. From the starting point, draw horizontal line to the right until it crosses the graph again (if it does not, the game is over). Then move the point $\pi/\omega$ to the left, adjust the vertical coordinate so that it lies on the graph of $F$, and repeat. After a few tries you should get a pretty good idea of how this dynamical system behaves.
(Old answer)
First of all, you can divide both sides by c. So let's assume this was done, and our equation is $g\cos\omega t+b = \exp(-t)$. This means we are looking for intersections between the exponential curve and a cosine wave. In general, there may be many intersections, in fact there are infinitely many when $ g>b$. Are all of these solutions relevant to your task, or is there a preferred one?
If $g<b$, we have an upper and lower bounds for solutions, namely $-\log(b\pm g)$. The number $-\log b$ looks like a good starting point for Newton's method, which can indeed be written in the form of a series.
I might be able to say more if you can clarify a) what to do with nonunique solutions, and b) if you know anything about the relative size of the constants involved. For example, is b much larger than g? Etc.
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