Помогите решить:

2sin^2x-3sinx-2=0, промежуток [-П; П]

Пусть t = sinx, t ∈ [-1; 1]

2t² - 3t - 2 = 0

D = 9 + 2·2·4 = 25 = 5²

t₁ = (3 + 5) / 4 = 8/4 = 2 - посторонний корень

t₂ = (3 - 5) / 4 = - 1/2

Обратная замена:

sinx = - 1/2

x = (-1) ⁿ⁺¹π/6 + πn, n ∈ Z

-π ≤ (-1) ⁿ⁺¹π/6 + πn ≤ π, n ∈ Z |·6/π

-6 ≤ (-1) ⁿ⁺¹ + 6n ≤ 6, n ∈ Z

n = - 1

x₁ = (-1) ⁻¹⁺¹π/6 - π = π/6 - π/6 = - 5π/6

n = 0

x₂ = (-1) ¹π/6 = - π/6

n = 1

x₃ = (-1) ²π/6 + π = 7π/6 - уже не подходит