Помогите решить

4sin^4x-5sin²x+1=0

Решение

sin²x = t

4t² - 5t + 1 = 0

D = 25 - 4*4*1 = 9

t₁ = (5 - 3) / 8

t₁ = 1/4

t₂ = (5 + 3) / 8

t₂ = 1

1) sin²x = 1/4

а) sinx = - 1/2

x = (-1) ^n * arcsin (- 1/2) + πk, k ∈ Z

x₁ = (-1) ^ (n+1) * π/6 + πk, k ∈ Z

б) sinx = 1/2

x = (-1) ^n * arcsin (1/2) + πn, n ∈ Z

x₂ = (-1) ^n * π/6 + πn, n ∈ Z

2) sin²x = 1

а) sinx = - 1

x₃ = - π/2 + 2πm, m ∈ Z

б) sinx = 1

x₄ = π/2 + 2πl, l ∈ Z