Trigonometric equations solutions within a range I wanted to ask a question about solving the following trigonometric equation:
$$ \tan\left(\theta + \frac{\pi}{6}\right) = \tan (\pi - \theta)$$
in the interval $0 ≤ \theta ≤ \pi$
This is the following route I took. 
$$ \theta + \frac{\pi}{6} = \pi - \theta$$
$$ 2 \theta = \frac{5 \pi}{6}$$
$$ \theta = \frac{5 \pi}{12}$$
Now, using the interval above, I then clarified that there are no further solutions as the next solution would be $+ \pi$ so would be $\frac{17 \pi}{12}$ and hence out of this range. 
However, I apparently made a slight "error". According to a friend in the lecture hall today, there is another solution, $\frac{11 \pi}{12}$ as follows:
$$ 2 \theta = \frac{5 \pi}{6}, \frac{11 \pi}{6}$$
$$ \theta = \frac{5 \pi}{12}, \frac{11 \pi}{12}$$
because in the range $0 ≤ 2 \theta ≤ \pi$, the result $\frac{11 \pi}{6}$ is possible.
I'm confused. How can it be that when I solved for $\theta = \frac{5 \pi}{12}$ there were no more solutions available in the interval but using $2 \theta = \frac{5 \pi}{6}$ there were solutions possible?

The point is that you are adding the period $\pi$ to $\theta$ but it is wrong indeed the following holds
$$\tan x=\tan y \iff x=y+k\pi$$
and thus we have that
$$\tan\left(\theta + \frac{\pi}{6}\right) = \tan \left(\pi - \theta\right)\iff \theta + \frac{\pi}{6} = \pi - \theta+k\pi$$
$$\iff 2\theta=\frac56\pi+k\pi\iff\theta=\frac5{12}\pi+k\frac \pi 2$$
therefore


*

*$\theta_1=\frac5{12}\pi$

*$\theta_2=\frac5{12}\pi+\frac \pi 2=\frac {11} {12}\pi$

