# Solve: tan2x + tan x = 0

• Apr 23rd 2009, 09:00 AM
SarahGr
Solve: tan2x + tan x = 0
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• Apr 23rd 2009, 09:09 AM
running-gag
Hi

You can use the trig identity $\displaystyle \tan(2x) = \frac{2\tan x}{1-tan^2x}$ and factor
• Apr 23rd 2009, 11:58 AM
masters
Quote:

Originally Posted by SarahGr
Solve: tan2x + tan x = 0

I have no idea how to solve this one =[

Hi SarahGr,

Solve:

$\displaystyle \tan 2x + \tan x = 0$

Recall that $\displaystyle \tan 2x=\frac{2 \tan x}{1-\tan^2 x}$

$\displaystyle \frac{2 \tan x}{1-\tan^2 x}+ \tan x=0$

$\displaystyle 2 \tan x+\tan x(1-\tan^2 x)=0$

$\displaystyle 2 \tan x+ \tan x - \tan^3 x=0$

$\displaystyle 3 \tan x - \tan^3 x=0$

$\displaystyle \tan x(3-\tan^2 x)=0$

$\displaystyle 3-\tan^2 x=0$

$\displaystyle \tan^2 x=3$

$\displaystyle \tan x = \sqrt{3}$

$\displaystyle x = 60^{\circ} \ \ or \ \ \frac{\pi}{3}$ radians and $\displaystyle 240^{\circ} \ \ or \ \ \frac{4\pi}{3}$ radians.
• Apr 23rd 2009, 12:18 PM
running-gag
$\displaystyle \tan^2 x=3$ can also lead to $\displaystyle \tan x = -\sqrt{3}$

$\displaystyle x = 120^{\circ}$

$\displaystyle x = 300^{\circ}$
• Apr 23rd 2009, 12:51 PM
e^(i*pi)
Quote:

Originally Posted by running-gag
$\displaystyle \tan^2 x=3$ can also lead to $\displaystyle \tan x = -\sqrt{3}$

$\displaystyle x = 120^{\circ}$

$\displaystyle x = 300^{\circ}$

$\displaystyle tan(x) = 0$ is also a solution

$\displaystyle x = 0^o \text { , } 0 \text { radians }$

$\displaystyle x = 180^o \text { , } \pi \text { radians }$

$\displaystyle x = 360^{\circ} \text { , } 2\pi \text { radians }$
• Apr 23rd 2009, 12:54 PM
running-gag
Quote:

Originally Posted by e^(i*pi)
$\displaystyle tan(x) = 0$ is also a solution

Of course it is ! (Happy)
• Apr 23rd 2009, 03:07 PM
SarahGr
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• Feb 5th 2013, 06:41 AM
scottsophie
Re: Solve: tan2x + tan x = 0
overall, with tan(x)=0, tan(x)=√3, tan(x)=-√3

between 0≤x≤360: x= 0°, 60°, 120°, 180°, 240°, 300°, 360°
• Feb 5th 2013, 07:34 AM
Petrus
Re: Solve: tan2x + tan x = 0
Quote:

Originally Posted by masters
Hi SarahGr,

Solve:

$\displaystyle \tan 2x + \tan x = 0$

Recall that $\displaystyle \tan 2x=\frac{2 \tan x}{1-\tan^2 x}$

$\displaystyle \frac{2 \tan x}{1-\tan^2 x}+ \tan x=0$

$\displaystyle 2 \tan x+\tan x(1-\tan^2 x)=0$

$\displaystyle 2 \tan x+ \tan x - \tan^3 x=0$

$\displaystyle 3 \tan x - \tan^3 x=0$

$\displaystyle \tan x(3-\tan^2 x)=0$

$\displaystyle 3-\tan^2 x=0$

$\displaystyle \tan^2 x=3$

$\displaystyle \tan x = \sqrt{3}$

$\displaystyle x = 60^{\circ} \ \ or \ \ \frac{\pi}{3}$ radians and $\displaystyle 240^{\circ} \ \ or \ \ \frac{4\pi}{3}$ radians.

Hello masters!
There is one step i did not understand

$\displaystyle \frac{2 \tan x}{1-\tan^2 x}+ \tan x=0$ i did understand how u got that but ur next step i did not understand ( could not fined it on ur code when i try copy)

any possible that some1 could explain:)?
• Feb 5th 2013, 07:55 AM
scottsophie
Re: Solve: tan2x + tan x = 0
you have (tan2x/1-tan2x) + tanx = 0
You find a common denominator of 1-tan2x

this gives you: tan2x + (tanx)(1-tan2x) = 0

(this equation is all over '1-tan2x', but you can get rid of it by moving it to the other side of the '=' sign, and multiplying this by the 0).

You then expand the brackets to get: tan2x + tanx - tan3x = 0

You then simplify this to get: tan3x + tanx = 0

Hope this helps.
• Feb 5th 2013, 08:58 AM
Petrus
Re: Solve: tan2x + tan x = 0
Thanks:) got it:) ii thought so.. but was not 100% it was correct:)