By trigonometry, $$f(\theta)= \log (2)\left( \sin\theta+\cos(\pi/6-\theta)-\cos(\pi/6+\theta) \right)=\log(4)\sin\theta.$$
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.