Question

Помогите с уравнением

Answer 1

Решение
 {cos2x - 3cosx - 1 = 0
 {(2sinx - 1) ≠ 0

cos2x - 3cosx - 1 = 0
2cos²x - 1 - 3cosx - 1 = 0
2cos²x - 3cosx - 2 = 0
cosx = t, I t I ≤ 1
2t² - 3t - 2 = 0
D = 9 + 4*2*2 = 25
t₁ = (3 - 5)/4 = - 1/2
t₂ = (3 + 5)/4 = 2 ∉ [-1;1]
cosx = 1/2
x = +-arccos(1/2) + 2πk, k ∈ Z
x = +-(π/3) + 2πk, k ∈ Z
2)√(2sinx - 1) ≠ 0
2sinx - 1 > 0
sinx > 1/2
arcsin(1/2) + 2πn < x < π - arcsin(1/2) + 2πn, n ∈ Z
π/6 + 2πn < x < π - π/6 + 2πn, n ∈ Z
π/6 + 2πn < x < 5π/6 + 2πn, n ∈ Z