How do you solve ${tan}^{2} x = \frac{1}{3}$ $tan^2x=1/3$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve ${tan}^{2} x = \frac{1}{3}$ $tan^2x=1/3$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2018-04-26T08:10:03

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6} and \frac{11 π}{6}, w h e r e, x \in [0, 2 π)$ $x=pi/6,(5pi)/6,(7pi)/6and(11pi)/6,where,x in [0,2pi)$

Explanation:

Here,

${tan}^{2} x = \frac{1}{3} = {(\frac{1}{\sqrt{3}})}^{2}, w h e r e, 0 \leq x \leq 2 π$ $tan^2x=1/3=(1/sqrt3)^2, where, 0 <= x <= 2pi$

$\Rightarrow tan x = \pm \frac{1}{\sqrt{3}}$ $=>tanx=+-1/sqrt3$

Now,

$(i) tan x = \frac{1}{\sqrt{3}} > 0 \Rightarrow I^{s t} Q u a d r a n t or I I I^{r d} Q u a d r a n t$ $(i) tanx=1/sqrt3 > 0=>I^(st)Quadrant or III^(rd)Quadrant$

$:.x=pi/6to I^(st)Quadrant or$

$x=pi+pi/6=(7pi)/6toIII^(rd) Quadrant$

$(ii)tanx=-1/sqrt3 < 0=>II^(nd)Quadrant or IV^(th)Quadrant$

$:.x=pi-pi/6=(5pi)/6toII^(nd)Quadrant or$

$x=2pi-pi/6=(11pi)/6to IV^(th)Quadrant$

Hence,

$x=pi/6,(5pi)/6,(7pi)/6and(11pi)/6$

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How do you solve ${tan}^{2} x = \frac{1}{3}$ $tan^2x=1/3$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve ${tan}^{2} x = \frac{1}{3}$ $tan^2x=1/3$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?