How do you solve ${(2Tan(x-pi/12)) / (1-Tan^2(x-pi/12))}^2=3$ from $[-pi/2, pi/2]$ ?

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Answer 1 · 2016-10-07T10:01:29

$x = -pi/4, -pi/12 and 5/12pi.$

Use #tan (2a) = (2 tan a )/(1-tan^2a).

Here, $tan^2 (2x-pi/6) = 3$ . So,

$tan(2x-pi/6)=+-sqrt3$

For the principal value of (2x-pi/6) in (-pi/2, pi/2),

$2x-pi/6=+-pi/3$ that gives

$x = 5/12pi and x = -pi/4$ .

The general value is given by.,

. $2x-pi/6=kpi+(+-pi/3), k=0.+-1.+-2,+-3,...$ that gives

$x = k/2pi+(pi/12+-pi/3), k=0.+-1.+-2,+-3,...$

$k = -1$ throws $-pi/12$ into $(-pi/2, pi/2)$

So, the answer is $x = -pi/12, -pi/4 and 5/12pi.$

How do you solve {(2Tan(x-pi/12)) / (1-Tan^2(x-pi/12))}^2=3{(2Tan(x-pi/12)) / (1-Tan^2(x-pi/12))}^2=3 from [-pi/2, pi/2][-pi/2, pi/2]?