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!
Supposing that $\displaystyle \tan x \ne 0$ You can devide both terms by $\displaystyle \tan x$ and after some steps You arrive at the equation $\displaystyle \tan x = \pm \sqrt{2}$ which has four solutions, one for each quadrant...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle 2cotx-tanx=0$
$\displaystyle \displaystyle\frac{2}{tanx}-tanx=0\Rightarrow\frac{2tanx}{tanx}-tan^2x=tanx(0)=0$
$\displaystyle 2-tan^2x=0\Rightarrow\ (\sqrt{2}+tanx)(\sqrt{2}-tanx=0)$
$\displaystyle tanx$ 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.
$\displaystyle x=tan^{-1}\sqrt{2},\;\;{\pi}+tan^{-1}\sqrt{2},\;\;2{\pi}-tan^{-1}\sqrt{2},\;\;{\pi}-tan^{-1}\sqrt{2}$