How do you solve ${tan}^{2} 2 x - 1 = 0$ $tan^2 2x -1=0$ ?

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How do you solve ${tan}^{2} 2 x - 1 = 0$ $tan^2 2x -1=0$ ?

1 Answer

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Answer 1 · 2016-10-23T23:53:42

$\frac{π}{8} + \frac{k π}{2}$ $pi/8 + (kpi)/2$
$\frac{7 π}{8} + \frac{k π}{2}$ $(7pi)/8 + (kpi)/2$

Explanation:

${tan}^{2} 2 x = 1$ $tan^2 2x = 1$
$tan 2 x = \pm 1$ $tan 2x = +- 1$
Trig Table and unit circle -->

a. tan 2x = 1
$2 x = \frac{π}{4} + k π$ $2x = pi/4 + kpi$
$x = \frac{π}{8} + \frac{k π}{2}$ $x = pi/8 + (kpi)/2$

b. tan 2x = - 1
$2 x = - \frac{π}{4} + k π$ $2x = -pi/4 + kpi$ , or $2 x = \frac{7 π}{4} + k π$ $2x = (7pi)/4 + kpi$ --> (co-terminal arcs)
$x = \frac{7 π}{8} + \frac{k π}{2}$ $x = (7pi)/8 + (kpi)/2$

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How do you solve ${tan}^{2} 2 x - 1 = 0$ $tan^2 2x -1=0$ ?

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