помогите решить
4sin^4x-5sin²x+1=0

Решениеsin²x = t4t² - 5t + 1 = 0D = 25 - 4*4*1 = 9t₁ = (5 - 3)/8t₁ = 1/4t₂ = (5 + 3)/8t₂ = 11)  sin²x = 1/4а) sinx = - 1/2x = (-1)^n * arcsin(- 1/2) + πk, k ∈ Zx₁ = (-1)^(n+1) * π/6 + πk, k ∈ Zб) sinx = 1/2x = (-1)^n * arcsin(1/2) + πn, n ∈ Zx₂ = (-1)^n * π/6 + πn, n ∈ Z2)  sin²x =  1а)  sinx = - 1x₃ = - π/2 + 2πm, m ∈ Zб)  sinx = 1x₄ = π/2 + 2πl, l ∈ Z