How do you solve $6 {tan}^{2} x - 2 = 4$ $6tan^2x-2=4$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-09-07T20:47:02

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

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

$\to 6 {tan}^{2} (x) = 6$ $rarr 6tan^2(x)=6$

$\to {tan}^{2} (x) = 1$ $rarr tan^2(x)=1$

$\to tan (x) = \pm 1$ $rarr tan(x)= +-1$

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=46tan2x−2=46tan^2x-2=4 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?