How do you find all solutions in [0, 2π): $3 {tan}^{2} x - 1 = 0$ $3tan^2x - 1 = 0$ ?

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How do you find all solutions in [0, 2π): $3 {tan}^{2} x - 1 = 0$ $3tan^2x - 1 = 0$ ?

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Answer 1 · 2016-10-11T10:30:55

Solution: In $[0, 2 π) x = \frac{π}{6}, x = \frac{7 π}{6} and x = \frac{5 π}{6}, x = \frac{11 π}{6}$ $[0,2pi) x=pi/6,x=(7pi)/6 and x=(5pi)/6, x= (11pi)/6$

Explanation:

$3 {tan}^{2} x - 1 = 0 or 3 {tan}^{2} x = 1 or {tan}^{2} x = \frac{1}{3} or tan x = \pm \frac{1}{\sqrt{3}}$ $3tan^2x-1=0 or3tan^2x=1 or tan^2x=1/3 or tanx=+- 1/sqrt3$ . We know $tan (\frac{π}{6}) = \frac{1}{\sqrt{3}} and tan (π + \frac{π}{6}) = \frac{1}{\sqrt{3}}$ $tan(pi/6)= 1/sqrt3 and tan(pi+pi/6)= 1/sqrt3$ and also
$tan (2 π - \frac{π}{6}) = - \frac{1}{\sqrt{3}} and tan (π - \frac{π}{6}) = - \frac{1}{\sqrt{3}}$ $tan(2pi-pi/6)= -1/sqrt3 and tan(pi-pi/6)= -1/sqrt3$ .
Solution: In $[0, 2 π) x = \frac{π}{6}, x = \frac{7 π}{6} and x = \frac{5 π}{6}, x = \frac{11 π}{6}$ $[0,2pi) x=pi/6,x=(7pi)/6 and x=(5pi)/6, x= (11pi)/6$ [Ans]

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How do you find all solutions in [0, 2π): $3 {tan}^{2} x - 1 = 0$ $3tan^2x - 1 = 0$ ?

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How do you find all solutions in [0, 2π): $3 {tan}^{2} x - 1 = 0$ $3tan^2x - 1 = 0$ ?