How do you solve the following equation ${tan}^{2} x = \frac{1}{3}$ $tan^2 x = 1/3$ in the interval [0, 2pi]?

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How do you solve the following equation ${tan}^{2} x = \frac{1}{3}$ $tan^2 x = 1/3$ in the interval [0, 2pi]?

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Answer 1 · 2016-10-31T13:19:30

$\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$ $pi/6, (5pi)/6, (7pi)/6, (11pi)/6$

Explanation:

${tan}^{2} x = \frac{1}{3}$ $tan^2 x = 1/3$
$tan x = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$ $tan x = +- 1/sqrt3 = +- sqrt3/3$
Use unit circle and trig table -->
a. $tan x = \frac{\sqrt{3}}{3}$ $tan x = sqrt3/3$ --> $x = \frac{π}{6}$ $x = pi/6$ , and $x = \frac{π}{6} + π = \frac{7 π}{6}$ $x = pi/6 + pi = (7pi)/6$
b. $tan x = - \frac{\sqrt{3}}{3}$ $tan x = -sqrt3/3$ --> $x = \frac{5 π}{6}$ $x = (5pi)/6$ and $x = \frac{5 π}{6} + π = \frac{11 π}{6}$ $x = (5pi)/6 + pi = (11pi)/6$

Answers for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$ $pi/6, (5pi)/6, (7pi)/6, (11pi)/6$

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How do you solve the following equation ${tan}^{2} x = \frac{1}{3}$ $tan^2 x = 1/3$ in the interval [0, 2pi]?

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How do you solve the following equation ${tan}^{2} x = \frac{1}{3}$ $tan^2 x = 1/3$ in the interval [0, 2pi]?