Question

4 tg ^2x+3/cos^2x+3= 0

Answer 1

4 *sin²x/ cos² x+ 3/cosx + 3 = 0     | * cos x
4 * sin² x + 3 * cos x + 3 * cos² x = 0
-cos² x + 3 * cos x + 4 = 0     | * (-1)
cos² x - 3 cos x - 4 = 0
cos x = t , t ∈ [-1 ; 1]
t² - 3t - 4 = 0
t₁ + t₂ = 3     t₁ = 4 , не подходит, т. к. t ∈ [-1 ; 1]
t₁ * t₂ = -4     t₂ = -1
cos x = -1
x = π + 2πn, n ∈ z