Question

Its PreCal I’ll give brainliest

Answer 1

Answer:

tan²x + tan x - 12 = 0 (1)

suppose: tan x = t

(1)=> t² + t - 12 = 0

a = 1    b = 1     c = -12

⇒ Δ = b² - 4ac = 1 + 48 = 49 > 0 => has 2 solutions

=> t = $\frac{1+\sqrt{49} }{2}=\frac{8}{2}=4$

or t = $\frac{1-\sqrt{49} }{2}=\frac{-6}{2}= -3$

because t = tan x => tan x = 4 or tan x = -3

because x ∈ [0;2pi) => with tan x = 4 => x = arctan(4)  or x = arctan(4) + pi

                                     with tan x = -3 => x = arctan(-3) + pi

Step-by-step explanation: