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 = [tex]\frac{1+\sqrt{49} }{2}=\frac{8}{2}=4[/tex]
or t = [tex]\frac{1-\sqrt{49} }{2}=\frac{-6}{2}= -3[/tex]
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:
