Как решить 3sinx - 2cos^x = -3?
3sinx - 2cos²x = -3 3sinx - 2(1-sin²x) = -3 3sinx - 2 + 2sin²x = -3 2sin²x + 3sinx + 1 = 0 sinx = t 2t² + 3t + 1 = 0 D = 9-8 = 1 √D = 1 x₁ = (-3-1)/4 = -1 x₂ = (-3+1)/4 = -1/2 sinx = -1 sinx = -1/2 x = -π/2 + 2πn x = -π/6 + 2πn x = -5π/6 + 2πn n∈Z