Решите уравнение 2cos x - cos 2x - cos^2x - вопрос №2227451888 от Ibrolya 13.09.2022 13:48

Question

Алгебра: Решите уравнение 2cos x - cos 2x - cos^2x

Ibrolya · Answer

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 = -1/3cosx = -1/3 -- x₁=arccos(-1/3) +2πn = arccos(1/3) +  2πn; x₂= π-arccos(-1/3) +2πncosx = 1 -- X=π/2+πn n∈Zответ: π/2+2πn, arccos(1/3) +  2πn, π - arccos(1/3) +  2πn; n∈Z...