Let us define function $f:R \rightarrow R$ as follows:
\begin{align} f(x)\triangleq \sum_{i=1}^K \frac{\alpha_i \gamma_i \sin(x-\theta_i)}{1+\gamma_i[1+\cos(x-\theta_i) ]}, \end{align} where anything except $x$ is a given parameter. I am trying to find the closed form solutions of $f(x)=0$. Does anybody know how to do that? If it is not possible to find a closed form solution, can we say something about the number of solutions? Is there finite number of solutions?
Any help would be appreciated.
Edit 1: Since $f(x)$ is periodic, I am only interested in finding the solutions located in the interval $[0~2\pi]$.
Edit 2: Is there a way to find every solution in $[0~2\pi]$ numerically?
Except in the case $K=1$, I do not see a way to obtain a closed form solution. As for the number of solutions, observe that (here I assume that $\gamma_i>-1$, so that the denominators never vanish) $$ f(x)=-\frac{d}{dx}\,\sum_{i=1}^K\log\bigl(1+\gamma_i(1+\cos(x-\theta_i))\bigr). $$ The roots of $f$ are precisely the critical points of $\sum_{i=1}^K\log\bigl(1+\gamma_i(1+\cos(x-\theta_i))\bigr)$. Since this is a smooth function of period $2\pi$, it attains its maximum and minimum on any interval of length $2\,\pi$, so that $f$ has at least two roots on every interval of length $2\,\pi$.