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.#