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