how do you solve: √3 tan^2x - (1+√3) tanx + 1 = 0
let v = tan x
then what you have is:
√3 v^2 - (1+√3) V + 1 = 0
let a = √3
now it is
a v^2 - (1+a) v + 1 = 0
looks like a plain old quadratic
v = { (1+a) +/- sqrt ( [a+1]^2 - 4 a) } / 2 a
v = { (1+a) +/- sqrt ( a^2 -2 a + 1) } / 2 a
v = { (a+1) +/- (a-1) } / 2 a
v = 1 or v = 1/a
tan x = 1 or tan x = 1/ (2 √3)
first quadrant and third quadrant solutions
check my arithmetic
√3 tan^2x - (1+√3) tanx + 1 = 0
so we have
tanx = ((1+√3)±√((1+√3)^2 - 4√3)) / (2√3)
= ((1+√3)±√(4+2√3 - 4√3)) / (2√3)
= ((1+√3)±√((1-√3)^2) / (2√3)
= ((1+√3)±(1-√3)) / (2√3)
= (1+√3+1-√3)/(2√3) or (1+√3-1+√3)/(2√3)
= 1/√3 or 1
x = π/4 + kπ
or π/6 + kπ