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