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There is this trig equation:

$$ 5\tan x - 2\tan 2x = 0 \text{ for 0 < 0 < 360 } $$

So far I've gotten $$\tan x = \text{0, 180}$$ and all I have to solve now is $$\tan ^2x = 0.2$$ which gives me two angle, $24.1$ and $-24.1$. For some reason, this is wrong? Can someone please tell me where I'm going wrong? Thank you in advance!

UPDATE I'm so sorry! I made a mistake in my mathjax, I fixed it.

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Hint : \begin{align} 5\tan x - 2\tan ^2x &= 0\\ (5 - 2\tan x)\tan x&=0\\ \tan x=0\qquad&;\qquad5 - 2\tan x=0 \end{align} – Tunk-Fey Jul 27 '14 at 12:38
@Shabbeh I made a mistake in my mathjax, so sorry! I fixed it though – Samir Chahine Jul 27 '14 at 12:43
It should be $\tan x = 2.5$ – Darth Geek Jul 27 '14 at 12:43
up vote 2 down vote accepted

Answer of the Original Version:

We have$$\tan x(2\tan x-5)=0$$

If $\tan x=0, x=n180^\circ$ where $n$ is any integer

If $\displaystyle2\tan x-5=0,\tan x=\frac52$

Google says $\displaystyle\arctan\frac52\approx68.1985905^\circ$

$\displaystyle\implies x\approx m180^\circ+68.1985905^\circ$ where $m$ is any integer

Probably you have meant $0<x<360^\circ\implies 0<m180^\circ+68.1985905^\circ<360^\circ$

Answer to the Edited Version:

Using Double Angle formula,

$$5\tan x=2\tan2x=2\frac{2\tan x}{1-\tan^2x}$$

$$\tan x(1-5\tan^2 x)=0$$

If $\tan x=0$ has been dealt already

$\displaystyle1-5\tan^2x=0\iff \tan^2x=\frac15\implies\cos2x=\frac{1-\tan^2x}{1+\tan^2x}=\frac23$

Using Google, $\displaystyle1\arccos\frac23\approx48.1896851$

$\displaystyle\implies2x=2m\pi\pm48.1896851^\circ\implies x=?$

Find $m$ such that $0<x<360^\circ$

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I'm sorry I made a mistake while writing out my question, I had a different question, sorry. – Samir Chahine Jul 27 '14 at 12:43
@SamirChahine, Please find the edited version – lab bhattacharjee Jul 27 '14 at 12:58

Answered to original questions:

$$5\tan x - 2\tan^2 x = 0 \iff \tan x(5 - 2\tan x) = 0$$

Than means $5\tan x = 0\iff \tan x = 0\;$ or $\;2\tan x = 5 \iff \tan x = \frac 52$.

Can you take it from here?

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Rectify the sign – lab bhattacharjee Jul 27 '14 at 12:42
I had an error while writing my question, sorry! I fixed my question but I'm sure this would have been correct otherwise. – Samir Chahine Jul 27 '14 at 12:44
@amWhy Why have you removed the "you" from "can you take it from here?"? – alexqwx Jul 27 '14 at 12:56
@alexqwx I thought 'you', I just forgot to write "you"! Thanks for pointing out the missing word! – amWhy Jul 27 '14 at 12:57

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