2sin²x/2 - 3√3sin x/2 + 3 = 0
Замена: sin x/2 = t, t ∈ [-1; 1]
2t² - 3√3t + 3 = 0
D = 27 - 24 = 3
t₁ = (3√3-√3)/4 = √3/2
t₂ = (3√3+√3)/4 = √3 -- не удовлетворяет ограничению t ∈ [-1; 1]
sin x/2 = √3/2
x/2 = (-1)ⁿ · π/3 + πn, n ∈ Z
x = (-1)ⁿ · 2π/3 + 2πn, n ∈ Z