Question

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

Answer 1

2cosx - (cos²x-sin²x) -cos²x = 0
2cosx - cos²x+sin²x - cos²x = 0
sin²x + 2cosx - 2cos²x = 0
1-cos²x + 2cosx - 2cos²x = 0
-3cos²x + 2cosx +1 = 0
cosx = t
-3t²+2t+1 = 0
D = 4 + 4*3 =16 = 4²
x₁ = (-2 - 4)/-6 =1
x₂ = -1/3
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cosx = -1/3 --> x₁=arccos(-1/3) +2πn = arccos(1/3) +  2πn; x₂= π-arccos(-1/3) +2πn

cosx = 1 --> X=π/2+πn n∈Z
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ответ: π/2+2πn, arccos(1/3) +  2πn, π - arccos(1/3) +  2πn; n∈Z

Answer 2

Спс