[SOLVED] A cube in a trig equation?

**DannyMath** · April 10th 2010, 08:32 AM

3tan^3x = tan x

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

**maddas** · April 10th 2010, 08:40 AM

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

**DannyMath** · April 10th 2010, 08:49 AM

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?

**maddas** · April 10th 2010, 08:51 AM

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

edit: yes.

**DannyMath** · April 10th 2010, 08:54 AM

Ok thank you!

**Prove It** · April 10th 2010, 09:08 AM

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 $n \in \mathbf{Z}$ .

**DannyMath** · April 10th 2010, 09:14 AM

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!

**DannyMath** · April 10th 2010, 09:31 AM

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 $n \in \mathbf{Z}$ ?

**Prove It** · April 11th 2010, 01:53 AM

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 $n \in \mathbf{Z}$ .

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