Hi, so I was doing this question and I found the solution $\theta = \frac{5}{12}\pi$ equating $\theta+\frac{\pi}6$ to $\pi-\theta$. However, there is also another solution for this equation $\frac{11}{12}\pi$, I don't understand how am I suppose to ensure that all solutions in a given range have been found.
We have: $$\tan(\pi-\theta)=\tan\left(\theta+\frac{\pi}{6}\right)$$ What you did is from this, conclude that $\pi-\theta=\theta+\frac{\pi}{6}$. However, this is false. The following is true: $$(\pi-\theta)-\left(\theta+\frac{\pi}{6}\right)=\pi n \text{ for some } n \in \Bbb{Z}$$ Now, we need to solve for $\theta$ in terms of $n$ to get all of the solutions. Simplify the left side: $$-2\theta+\frac{5\pi}{6}=\pi n \text{ for some } n \in \Bbb{Z}$$ Solve for $\theta$: $$\theta=\frac{5\pi}{12}-\frac{\pi}{2} n \text{ for some } n \in \Bbb{Z}$$ Now, here are some of the solutions: $$n=1 \implies \theta=-\frac{\pi}{12}$$ $$n=0 \implies \theta=\frac{5\pi}{12}$$ $$n=-1 \implies \theta=\frac{11\pi}{12}$$ $$n=-2 \implies \theta=\frac{17\pi}{12}$$ If we are talking about the range of $\theta \in [0, \pi]$, $n=0$ and $n=-1$ are the only solutions since $n=1$ and $n=-2$ are outside of this range. However, for $\theta \in \Bbb{R}$, there are an infinite number of solutions.