question about a trig equation

**iiharthero** · June 19th 2009, 10:53 PM

could somebody please help me with equation ?
i think ive just made a stupid mistake because its not working T.T

Write general solution

tan2x = cot x

tan 2x = 2tanx/(1-tansquaredx)

therefore

2tanx/(1-tansquaredx) = 1/tanx

2tansquared x = 1/2

tan squared x = 1/4

tan x = + or - 1/2

which doesnt work
and the answers say the solution is npi + or - (pi/6)

thanks for any help !!

**malaygoel** · June 19th 2009, 11:00 PM

Originally Posted by iiharthero

2tanx/(1-tansquaredx) = 1/tanx

2tansquared x = 1/2................WRONG

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

$2tan^2x=1-tan^2x$

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

**yeongil** · June 19th 2009, 11:04 PM

$\begin{aligned} \tan 2x &= \cot x \\ \frac{2\tan x}{1 - \tan^2 x} &= \frac{1}{\tan x} \\ 2\tan^2 x &= 1 - \tan^2 x \\ 3\tan^2 x &= 1 \\ \tan^2 x &= \frac{1}{3} \\ \tan x &= \frac{\sqrt {3}}{3} \end{aligned}$
$x = \frac{\pi}{6} \pm n\pi$

01

EDIT: Beaten to it by malaygoel!

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