How do you solve ${tan}^{2} x + tan x = \sqrt{3} tan x + \sqrt{3}$ $tan^2x+tanx=sqrt3 tanx + sqrt3$ for the domain $0 \leq x \leq 360$ $0<=x<=360$ ?

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Answer 1 · 2018-06-23T13:29:26

${tan}^{2} x + tan x = \sqrt{3} tan x + \sqrt{3}$ $tan^2x+tanx=sqrt3 tanx + sqrt3$

$\Rightarrow {tan}^{2} x + tan x - \sqrt{3} tan x - \sqrt{3} = 0$ $=>tan^2x+tanx-sqrt3 tanx - sqrt3=0$

$\Rightarrow tan x (tan x + 1) - \sqrt{3} (tan x + 1) = 0$ $=>tanx(tanx+1)-sqrt3( tanx +1)=0$

$\Rightarrow (tan x + 1) (tan x - \sqrt{3}) = 0$ $=>(tanx+1)(tanx-sqrt3)=0$

When

#tanx+1=0

$\Rightarrow tan x = - 1 = tan (180 - 45) = tan (360 - 45)$ $=>tanx=-1=tan(180-45)=tan(360-45)$

So $x = 135^{\circ} or 315^{\circ}$ $x=135^@ or 315^@$

When

$tan x - \sqrt{3} = 0$ $tanx-sqrt3=0$

$\Rightarrow tan x = \sqrt{3} = {tan 60}^{\circ} = tan (180 + 60)$ $=>tanx=sqrt3=tan60^@=tan(180+60)$

So $x = 60^{\circ} or 240^{\circ}$ $x=60^@ or 240^@$

How do you solve tan2x+tanx=√3tanx+√3tan^2x+tanx=sqrt3 tanx + sqrt3 for the domain 0≤x≤3600<=x<=360?