Help Needed with this Trig Equation!

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February 3rd 2011, 08:16 PM
sammysparrow

Help Needed with this Trig Equation!

Hi, could someone show me how this one's done?

Find all solutions for X, satisfying 0 < X < (two pi):
2cotX = tanX

Thanks, your help would be much appreciated!
February 3rd 2011, 08:30 PM
chisigma

Supposing that $\tan x \ne 0$ You can devide both terms by $\tan x$ and after some steps You arrive at the equation $\tan x = \pm \sqrt{2}$ which has four solutions, one for each quadrant...

Kind regards

$\chi$ $\sigma$
February 3rd 2011, 08:44 PM
sammysparrow

Thanks very much!
February 4th 2011, 04:58 PM
Archie Meade

Quote:

Originally Posted by sammysparrow

Hi, could someone show me how this one's done?

Find all solutions for X, satisfying 0 < X < (two pi):
2cotX = tanX

Thanks, your help would be much appreciated!

$2cotx-tanx=0$

$\displaystyle\frac{2}{tanx}-tanx=0\Rightarrow\frac{2tanx}{tanx}-tan^2x=tanx(0)=0$

$2-tan^2x=0\Rightarrow\ (\sqrt{2}+tanx)(\sqrt{2}-tanx=0)$

$tanx$ gives the slope of a line passing through the origin with an angle x,

hence the one with positive slope traverses quadrants 1 and 3,
while the other traverses quadrants 2 and 4.

$x=tan^{-1}\sqrt{2},\;\;{\pi}+tan^{-1}\sqrt{2},\;\;2{\pi}-tan^{-1}\sqrt{2},\;\;{\pi}-tan^{-1}\sqrt{2}$
February 6th 2011, 06:02 PM
sammysparrow

Thanks very much, that's very helpful. (Clapping)