Let
$x\_s = \sin(\theta+\frac{2\pi s}{3})$ and
$y\_s = 1+\cos(\frac{2\pi s}{3})$, $s=0,1,2$.


Define $f(\theta) = \sum\_{s=0}^2 x\_s\ln y\_s$.


Is there any method to derive roots of $f(\theta)$.
I have run a simulation on it, and found that $\theta=0$ is a solution.
But I am unable to see how to analytically obtain it.



Roots of an entropy-like function
By trigonometry, $$f(\theta)= \log (2)\left( \sin\theta+\cos(\pi/6-\theta)-\cos(\pi/6+\theta) \right)=\log(4)\sin\theta.$$




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For the revised problem, we have this: $f(\theta)$ is $2\pi/3$-periodic, and $f(\theta)$ is odd, so it suffices to find the roots between $0$ and $\pi/3$ (both of which are themselves roots). Plotting indicates that $f(\theta)$ is unimodal on this interval, $f'(\theta)$ is strictly increasing, and $f''(\theta)$ is strictly increasing, and $f'''(\theta)\geq 6$. Each of these observations follows from the one after it (sometimes needing to also evaluate at $\theta=0$), and the last one seems easiest to prove. Not elegant, certainly, but it should get the job done.


