ПОМОГИТЕ СРОЧНО РЕШИТЬ


2in^2x-3 cosx+3=0

2Sin²x - 3Cosx + 3 = 02(1-Cos²x) - 3Cosx + 3 = 02 - 2Cos²x - 3Cosx + 3 = 0-2Cos²x - 3Cosx + 5 = 02Cos²x + 3Cosx - 5 = 0Cosx = t ∈ [-1;1]2t² + 3t - 5 = 0D = 9 - 4 * 2 * (-5) = 49t₁ = (-3 + 7) / 4 = 1t₂ = (-3 - 7) / 4 = -2.5 ∉ [-1;1]Cosx = 1x = 2πn, n∈Z