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