General solultion to $f(t) - f(t+{2\pi \over c})=A\sin(w_ot)$ While trying to solve the wave equation with non-homogeneous boundary conditions, I came across the final subproblem which involves finding the general solution to the following equation:
$$f(t) - f(t+{2\pi \over c})=A\sin(w_ot) \tag1$$
where $f(t)$ is the general solution, $t$ is time, and $c$, $w_o$ and $A$ are known constants. I used up all the space and time boundary conditions. The problem is that I don't know the general solution of equation $(1)$ or how to determine it.
The first thing that came to my mind was to expand $f(t+{2\pi\over c})$ into the first-order Taylor polynomial. However, That did not work because I do not have any more boundary conditions I can use. I was left with a constant that I could not determine.
So the second thing I tried was the Fourier transformation which gave me:
$$
F(jw)-e^{jw {2\pi\over c}}F(jw)= A{\pi\over j}\Big(\delta(w-w_o) - \delta(w+w_o)\Big) \tag2
$$
This allowed me to express $F(jw)$ as:
$$F(jw)= {A{\pi\over j}\Big(\delta(w-w_o) - \delta(w+w_o)\Big)} \cdot {1\over1-e^{jw {2\pi\over c}}} \tag3$$
and then to express $f(t)$ as:
$$f(t)=A\sin(\omega_o t) * \Phi^{-1} \Bigg\{ {1\over1-e^{jw {2\pi\over c}}} \Bigg\}  \tag4$$
where $*$ is the convolution operator and $\Phi^{-1} \{arg\}$ is the inverse Fourier transformation. It is ok for me if the convolution operator is in the expression of the general solution because Matlab can compute it. However, I was not able to solve the inverse Fourier transformation:
$$\Phi^{-1} \Bigg\{ {1\over1-e^{jw {2\pi\over c}}} \Bigg\}= {1\over 2\pi} \int _{-\infty}^{-\infty} { e^{jwt}\over1-e^{jw {2\pi\over c}}}dw\tag5$$
I tried using the integration by parts method and the substitution method but failed.
To summarize, I am trying to find the general solution of equation $(1)$. I ran out of boundary conditions I can use and failed to solve equation $(5)$ which would give me the solution I am looking for.
My questions are:
$1$. Is there a solution to equation $(5)$? If yes, what is it and how can I derive it?
$2$. If a general solution to equation $(1)$ exists, what is it and how can it be derived?
Thank you for your time.
 A: Note that for any test funtion $f$, we have $f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)$ in the sense of distributions.  Hence, beginning with the expression for $F(j\omega)$ (i.e., Equation $(2)$ in the posted question), we have in distribution
$$F(j\omega)=-j\pi A\left(\frac{\delta(\omega-\omega_0)}{1-e^{j2\pi \omega_0/c}}-\frac{\delta(\omega+\omega_0)}{1-e^{-j2\pi \omega_0/c}} \right)\tag1$$
Taking the inverse Fourier transform of $(1)$ yields
$$\begin{align}
f(t)&=-j\pi A \frac1{2\pi}\int_{-\infty}^\infty \left(\frac{\delta(\omega-\omega_0)}{1-e^{j2\pi \omega_0/c}}-\frac{\delta(\omega+\omega_0)}{1-e^{-j2\pi \omega_0/c}} \right) e^{j\omega t}\,d\omega\\\\
&=\frac{\pi A}{2\pi j}\left(\frac{e^{j\omega_0t}}{1-e^{j2\pi \omega_0 /c}}-\frac{e^{-j\omega_0t}}{1-e^{-j2\pi \omega_0 /c}}\right)\\\\
&=A\text{Im}\left(\frac{e^{j\omega_0t}}{1-e^{j2\pi \omega_0 /c}}\right)\\\\
&=A\text{Im}\left(\frac{je^{j\omega_0(t-\pi/c)}}{2\sin(.\pi \omega_0 /c))}\right)\\\\
&=\frac A2 \frac{\cos(\omega_0(t-\pi/c))}{\sin(\pi \omega_0/c)}
\end{align}$$

Solution Verification:
$$f(t)-f(t+\pi/c)=\frac A2\left(\frac{\cos(\omega_0(t-\pi/c))-\cos(\omega_0(t+\pi/c))}{\sin(\pi \omega_0/c)}\right)=A \sin(\omega_0 t)$$
as expected!
