
Trig Equation
This is a question taken from my grade 12 math textbook. I'm really not sure how to solve this problem, so any help is really appreciated!
Determine solutions for tanxcos^2x  tanx = 0 in the interval x Є [2p, 2p]
The answer taken from the textbook is: 2p,p, 0, p, 2p
Thanks in advance! :)

$\displaystyle \tan x \cos 2x  \tan x = 0$
$\displaystyle \tan x \cos 2x = \tan x $
divide both sides by tan?
$\displaystyle \cos 2x = 1$
Can you finish it?

$\displaystyle \displaystyle \tan{x}\cos^2{x}  \tan{x} = 0$
$\displaystyle \displaystyle \tan{x}(\cos^2{x}  1) = 0$
$\displaystyle \displaystyle \tan{x} = 0$ or $\displaystyle \displaystyle \cos^2{x}  1 = 0$
$\displaystyle \displaystyle \tan{x} = 0$ or $\displaystyle \displaystyle \cos{x} = \pm 1 $
$\displaystyle \displaystyle x = n\pi$ where $\displaystyle \displaystyle n \in \mathbf{Z}$ or $\displaystyle \displaystyle x = m\pi$ where $\displaystyle \displaystyle m \in \mathbf{Z}$.
Therefore in the period $\displaystyle \displaystyle [2\pi, 2\pi]$ the solutions are $\displaystyle \displaystyle \left\{2\pi, \pi, 0, \pi, 2\pi\right\}$.
BTW Pickslides, $\displaystyle \displaystyle \tan{x}$ could be $\displaystyle \displaystyle 0$, so you really shouldn't divide both sides by it...
