Решите уравнение: cos (2x+x) + 2cosx=0

Cos (2x+x) + 2cosx=0

cos2x cosx - sin2x sinx + 2cosx=0

(cos²x-sin²x) cosx - 2sinx cosx sinx + 2cosx=0

(cos²x-sin²x) cosx - 2sin²x cosx + 2cosx=0

cosx (cos²x-sin²x - 2sin²x + 2) = 0

cosx (cos²x-3sin²x+2) = 0

cosx=0 cos²x-3sin²x+2=0

x=π + πn cos²x-3 (1-cos²x) + 2=0

2 cos²x - 3 + 3cos²x+2=0

4cos²x-1=0

(2cosx-1) (2cosx+1) = 0

2cosx-1=0 2cosx+1=0

2cosx=1 2cosx=-1

cosx = 1 cosx = - 1

2 2

x = + arccos 1 + 2πn x = + arccos (-1) + 2πn

2 2

x = + π + 2πn x = + (π - π) + 2πn

3 3

x=+2π + 2πn

3

Ответ: х = π + πn

2

x = + π + 2πn

3

x=+2π + 2πn

3