Please take a look at this question attached, thank you.

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- Oct 15th 2008, 07:33 AMgearshifterSolve for X
Please take a look at this question attached, thank you.

- Oct 15th 2008, 11:40 AMShyamReply
$\displaystyle 3 \cot^2 x = 1$

$\displaystyle \cot^2 x = \frac{1}{3}$

$\displaystyle \cot x = \pm \frac{1}{\sqrt {3}}$

$\displaystyle \tan x = \pm \sqrt {3}$

$\displaystyle x = \frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}$

x = {$\displaystyle \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} $}