How do you solve $6tan^2x-2=4$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-09-07T20:47:02

$x in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}$

$6tan^2(x)-2=4$

$rarr 6tan^2(x)=6$

$rarr tan^2(x)=1$

$rarr tan(x)= +-1$

That is using a unit circle, the opposite and adjacent sides are of equal length, giving the 4 possibilities identified above.

How do you solve 6tan^2x-2=46tan^2x-2=4 for 0<=x<=2pi0<=x<=2pi?