quadratic equation help in my trig class needed

Question

solve for x where pi ≤ x ≤ 2pi.

tan^2 x + 2sqrt(3) tan x + 3 = 0.

Answer 1

First, replace "tan x" with y to get:

y² + 2√3 * y + 3 = 0

This factors to (y + √3)², which gives us y = -√3

replace y with tan x to get tan x = -√3

So---x = atan -√3 = -60 degrees or + 120 degrees (the 2 solutions that fall within 0 to 2pi)