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.