n∈Z
Пусть n = 2, n = 3, n = 4, n = 5
x₁ = (-1)² × π/6 - 5π/12 + π×2/2 = 1 × π/6 - 5π/12 + π = π/6 - 5π/12 + π = 2π/12 - 5π/12 + 12π/12 = 9π/12 = 3π/4
x₂ = (-1)³ × π/6 - 5π/12 + π×3/2 = -1 × π/6 - 5π/12 + 3π/2 = -π/6 - 5π/12 + 3π/2 = -2π/12 - 5π/12 + 18π/12 = 11π/12
x₃ = (-1)⁴ × π/6 - 5π/12 + π×4/2 = 1 × π/6 - 5π/12 + 4π/2 = π/6 - 5π/12 + 4π/2 = 2π/12 - 5π/12 + 24π/12 = 21π/12 = 7π/4
x₄ = (-1)⁵ × π/6 - 5π/12 + π×5/2 = -1 × π/6 - 5π/12 + 5π/2 = -π/6 - 5π/12 + 5π/2 = -2π/12 - 5π/12 + 30π/12 = 23π/12
ответ: x₁ = 3π/4, x₂ = 11π/12, x₃ = 7π/4, x₄ = 23π/12
Дискриминант отрицательный. Значит уравнение не имеет действительных корней.