# Finding closed form expression for the roots of $f(x) = \sum_{i=1}^K \frac{\alpha_i \gamma_i \sin(x-\theta_i)}{1+\gamma_i[1+\cos(x-\theta_i) ]}$

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.

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.
• Thanks. I am trying to digest your suggestion. Let's say K=2, hence we have $P(s,c)=A_s s+A_c c+A_{sc} sc+ A_{ss} s^2+ A_{cc} c^2$. Would you please tell me how I can find $R(s)$ now? I did not find a way to calculate the resultant. – James Apr 10 '18 at 19:15
• In this case $$R(s) = \left( {A_{{{\it cc}}}}^{2}-2\,A_{{{\it cc}}}A_{{{\it ss}}}+{A_{{{ \it sc}}}}^{2}+{A_{{{\it ss}}}}^{2} \right) {s}^{4}+ \left( 2\,A_{{c}} A_{{{\it sc}}}-2\,A_{{{\it cc}}}A_{{s}}+2\,A_{{s}}A_{{{\it ss}}} \right) {s}^{3}+ \left( {A_{{c}}}^{2}-2\,{A_{{{\it cc}}}}^{2}+2\,A_{{ {\it cc}}}A_{{{\it ss}}}+{A_{{s}}}^{2}-{A_{{{\it sc}}}}^{2} \right) {s }^{2}+ \left( -2\,A_{{c}}A_{{{\it sc}}}+2\,A_{{{\it cc}}}A_{{s}} \right) s-{A_{{c}}}^{2}+{A_{{{\it cc}}}}^{2}$$ Most computer algebra systems have a resultant function. – Robert Israel Apr 10 '18 at 22:49