# Thread: [SOLVED] A cube in a trig equation?

1. ## [SOLVED] A cube in a trig equation?

3tan^3x = tan x

I would like to solve this myself but any hint to what first step is sought

2. Let $u := \tan x$. You have $3u^3 = u$. One solution is $u=0$. If $u\neq0$, divide through by u to get $3u^2 = 1$ or $|u| = \frac{\sqrt{3}}3$. So you need the solutions to $\tan x = 0$ or $\tan x = \pm\frac{\sqrt{3}}3$.

3. Ok I see how you got sqrt(3)/3 answer, how do you go about getting 0?

Ok so setting u to 0 gets you (3)(0^2) = 0 which is true?

4. It should be clear that if $\tan x = 0$, then $3\tan^3x = \tan x$ since both sides would be zero...

edit: yes.

5. Ok thank you!

6. As an alternative:

$3\tan^3{x} = \tan{x}$

$3\tan^3{x} - \tan{x} = 0$

$\tan{x}(3\tan^2{x} - 1) = 0$

$\tan{x} = 0$ or $3\tan^2{x} - 1 = 0$.

Case 1: $\tan{x} = 0$

$x = \pi n$ where $n \in \mathbf{Z}$.

Case 2: $3\tan^2{x} - 1 = 0$

$3\tan^2{x} = 1$

$\tan^2{x} = \frac{1}{3}$

$\tan{x} = \pm \frac{1}{\sqrt{3}}$

$x = \left\{\frac{\pi}{6}, \pi - \frac{\pi}{6}, \pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6} \right\} + 2\pi n$ where $n \in \mathbf{Z}$

$x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\} + 2\pi n$.

Putting it all together gives:

$x = \left\{ 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{11\pi}{6} \right\} + 2\pi n$ where $n \in \mathbf{Z}$.

7. Wow man that's exactly how my book's been going about solving things, you just gotta bring it to the other side and factor it! Getting better by the day thank you!

8. Originally Posted by Prove It

Putting it all together gives:

$x = \left\{ 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{11\pi}{6} \right\} + 2\pi n$ where $n \in \mathbf{Z}$.
Could I also write it as:

$x = \pi n, \frac{\pi}{6} + \pi n, \frac{5\pi}{6} + \pi n$ where $n \in \mathbf{Z}$?

9. Yes you could.

Or if you wanted to use the set notation...

$\left\{0, \frac{\pi}{6}, \frac{5\pi}{6}\right\} + \pi n$ where $n \in \mathbf{Z}$.