2cos^2x=3sinx решите

2 (1-sin²x) - 3sin x = 0

2 - 2sin²x - 3sin x = 0

2sin²x + 3sin x - 2 = 0

Вводим замену sin x = t

2t² + 3t - 2 = 0

D = 9 + 16 = 25

t₁ = 1/2

t₂ = - 2

sin x = 1/2

x = (-1) ^n * π/6 + πn, n∈Z

sin x ≠ - 2

Ответ. x = (-1) ^n * π/6 + πn, n∈Z