# Solving trigonometric equations help

• Jan 12th 2008, 09:32 AM
mathgeek777
Solving trigonometric equations help
Hey guys.

Here's the problem. We did not cover a problem like this in class on Thursday, so that's why I'm confused.

$3\tan^22x-1=0$

I began to solve the trigonometric equation, and here's how I went about it.

$3\tan^22x = 1$

$tan^22x = \frac{1}{3}$

Here's where I'm stuck. I'm thinking of taking the square root of both sides, but then I'm left with:

$\tan2x = \pm\frac{\sqrt{3}}{3}$

And then once I divide two out, I'm left with a weird fraction that doesn't look like anything that can be found on the unit circle.

Am I doing anything wrong here?
• Jan 12th 2008, 10:09 AM
topsquark
Quote:

Originally Posted by mathgeek777
Hey guys.

Here's the problem. We did not cover a problem like this in class on Thursday, so that's why I'm confused.

$3\tan^22x-1=0$

I began to solve the trigonometric equation, and here's how I went about it.

$3\tan^22x = 1$

$tan^22x = \frac{1}{3}$

Here's where I'm stuck. I'm thinking of taking the square root of both sides, but then I'm left with:

$\tan2x = \pm\frac{\sqrt{3}}{3}$

And then once I divide two out, I'm left with a weird fraction that doesn't look like anything that can be found on the unit circle.

Am I doing anything wrong here?

Tsk tsk.

$tan^{-1} \left ( \frac{\sqrt{3}}{3} \right ) = \frac{\pi}{6}$

Now, this is a reference angle, and tangent is positive. Thus your possible values for 2x are $\frac{\pi}{6}, \frac{7\pi}{6}$.

So
$tan(2x) = \frac{\sqrt{3}}{2} \implies 2x = \frac{\pi}{6}, \frac{7\pi}{6}$

Now you do it for $tan^{-1} \left ( -\frac{\sqrt{3}}{3} \right )$.

-Dan
• Jan 12th 2008, 10:50 AM
mathgeek777
The weird thing is that I'm supposed to use this equation to verify that $x = \frac{\pi}{12}$ and $x=\frac{5\pi}{12}$ are solutions of the equation.
• Jan 12th 2008, 01:21 PM
topsquark
Quote:

Originally Posted by mathgeek777
The weird thing is that I'm supposed to use this equation to verify that $x = \frac{\pi}{12}$ and $x=\frac{5\pi}{12}$ are solutions of the equation.

Why is that weird? $\frac{\pi}{12}$ is a solution using $tan(2x) = +\frac{\sqrt{3}}{3}$ and $\frac{5\pi}{12}$ is a solution using $tan(2x) = -\frac{\sqrt{3}}{3}$.

-Dan
• Jan 12th 2008, 03:20 PM
mathgeek777
Quote:

Originally Posted by topsquark
Why is that weird? $\frac{\pi}{12}$ is a solution using $tan(2x) = +\frac{\sqrt{3}}{3}$ and $\frac{5\pi}{12}$ is a solution using $tan(2x) = -\frac{\sqrt{3}}{3}$.

-Dan

I'm just trying to figure out how you arrived at the answer.

EDIT: Nevermind, I just realized it.