Question

How to solve this for 0 ≤ x ≤ 2π ?

Answer 1

Given: `tan^2(x)+tan(x) = sqrt3tan(x) + sqrt3, 0 <= x < 2pi`

Add `-(sqrt3tan(x) + sqrt3)` to both sides:

`tan^2(x)+(1-sqrt3)tan(x) - sqrt3= 0, 0 <= x < 2pi`

Factor:

`(tan(x)+1)(tan(x) - sqrt3)= 0, 0 <= x < 2pi`

`tan(x) = -1` and `tan(x) = sqrt3, 0 <= x < 2pi`

`x = tan^-1(-1)` and `x = tan^-1(sqrt3), 0 <= x < 2pi`

`x = 3/4pi, 7/4pi, 1/3pi, and 4/3pi`

Answer 2

That's a quadratic equation in `\tan x.`

`tan ^2 x + \tan x = \sqrt{3} \tan x + \sqrt{3}`

`tan^2 x +(1-\sqrt{3}) \tanx - \sqrt{3} = 0`

Despite the square roots, this isn't too hard to factor:

`( \tan x - \sqrt{3} )(\tan x + 1) = 0`

`\tan x = -1` or `\tan x = \sqrt{3}`

My pet peeve is every problem in trig is 30,60,90 or 45,45,90; this one's both!

`x = frac{3\pi}{ 4}` or `x = frac{7 \pi}{4}` or `x=frac{pi}{4}` or `x=\frac{4\pi}{3}`