Решите уравнение
2cos x - cos 2x - cos^2x

Спс

2cosx - (cos²x-sin²x) -cos²x = 02cosx - cos²x+sin²x - cos²x = 0sin²x + 2cosx - 2cos²x = 01-cos²x + 2cosx - 2cos²x = 0-3cos²x + 2cosx +1 = 0cosx = t-3t²+2t+1 = 0D = 4 + 4*3 =16 = 4²x₁ = (-2 - 4)/-6 =1x₂ = -1/3------------------cosx = -1/3 --> x₁=arccos(-1/3) +2πn = arccos(1/3) +  2πn; x₂= π-arccos(-1/3) +2πncosx = 1 --> X=π/2+πn n∈Z---------------------------------------ответ: π/2+2πn, arccos(1/3) +  2πn, π - arccos(1/3) +  2πn; n∈Z