How do you solve ${tan}^{2} x - 3 = 0$ $tan^2x-3=0$ and find all solutions in the interval $0 \leq x < 360$ $0<=x<360$ ?

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Answer 1 · 2018-04-12T16:32:33

$x = 60^{\circ}, 120^{\circ}, 240^{\circ}, 300^{\circ} .$ $x=60^circ,120^circ,240^circ,300^circ.$

Here,

${tan}^{2} x - 3 = 0$ $tan^2x-3=0$

$\Rightarrow {tan}^{2} x = 3 = {(\sqrt{3})}^{2}$ $=>tan^2x=3=(sqrt3)^2$

$\Rightarrow tan x = \pm \sqrt{3}, w h e r e, x \in [0, 360)$ $=>tanx=+-sqrt3,where,x in [0,360)$

We have two option:

$(1) tan x = \sqrt{3} \Rightarrow x = 60^{\circ}, 240^{\circ}$ $(1)tanx=sqrt3=>x=60^circ,240^circ$

$(2) tan x = - \sqrt{3} \Rightarrow x = 120^{\circ}, 300^{\circ}$ $(2)tanx=-sqrt3=>x=120^circ,300^circ$

From, $(1) and (2)$ $(1)and(2)$

$x = 60^{\circ}, 120^{\circ}, 240^{\circ}, 300^{\circ} .$ $x=60^circ,120^circ,240^circ,300^circ.$

How do you solve tan^2x-3=0tan2x−3=0tan^2x-3=0 and find all solutions in the interval 0<=x<3600≤x<3600<=x<360?