Trigonometric equations solutions within a range

Question

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?

Answer 1

If $0 \le \theta \le \pi$, then $0 \le 2\theta \le 2\pi$. Since tangent is periodic with period $\pi$, you need to find all values in $[0,2\pi]$ for $2\theta$ to ensure that $\theta \in [0,\pi]$.

Answer 2

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

• $\theta_1=\frac5{12}\pi$
• $\theta_2=\frac5{12}\pi+\frac \pi 2=\frac {11} {12}\pi$