Trig Equation

**ZaraR** · November 28th 2010, 03:41 PM

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!

**pickslides** · November 28th 2010, 03:48 PM

$\tan x \cos 2x - \tan x = 0$

$\tan x \cos 2x = \tan x$

divide both sides by tan?

$\cos 2x = 1$

Can you finish it?

**Prove It** · November 28th 2010, 03:51 PM

$\displaystyle \tan{x}\cos^2{x} - \tan{x} = 0$

$\displaystyle \tan{x}(\cos^2{x} - 1) = 0$

$\displaystyle \tan{x} = 0$ or $\displaystyle \cos^2{x} - 1 = 0$

$\displaystyle \tan{x} = 0$ or $\displaystyle \cos{x} = \pm 1$

$\displaystyle x = n\pi$ where $\displaystyle n \in \mathbf{Z}$ or $\displaystyle x = m\pi$ where $\displaystyle m \in \mathbf{Z}$ .

Therefore in the period $\displaystyle [-2\pi, 2\pi]$ the solutions are $\displaystyle \left\{-2\pi, -\pi, 0, \pi, 2\pi\right\}$ .

BTW Pickslides, $\displaystyle \tan{x}$ could be $\displaystyle 0$ , so you really shouldn't divide both sides by it...

**ZaraR** · November 29th 2010, 12:27 PM

Ahh, okay. Thank you!

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