Let us define function $f:[0~ 2\pi] \rightarrow R$ 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 and we have $\alpha\_i >0, \forall i$ and $\gamma\_i >0, \forall i$. I am trying to find the closed form solutions of $f(x)=0$ in interval $[0~2\pi]$. 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.



Finding closed form expression for the roots of $f(x) = \sum_{i=1}^K \frac{\alpha_i \gamma_i \sin(x-\theta_i)}{1+\gamma_i[1+\cos(x-\theta_i) ]}$
Even for the case $K=2$, closed form solutions seem hopeless. Counting the number of solutions should be possible, though. Expand the sines and cosines and put everything over a common denominator: the numerator will be a trigonometric polynomial $P(\sin(x), \cos(x))$. Let $R(s)$ be the resultant of $P(s,c)$ and $s^2 + c^2-1$ with respect to $c$. This is a polynomial in $s$ whose roots are the values of $s$ where $P(s,c)$ and $s^2+c^2-1$ have a common solution. Use 
Sturm's theorem to count the number of such $s$ in $[-1,1]$. This may not be quite the number of solutions, because a given value of $\sin(x)$ may correspond to either one or two solutions with $\cos(x) = \pm \sqrt{1-\sin^2(x)}$, but that
shouldn't be too hard to handle.


