Question

помогите пожалуйста)
cos2x+cos^x+2cosx=0

Answer 1

cos2x + cos²x + 2cosx = 0
cos²x - sin²x + cos²x + 2cosx = 0
cos²x - (1-cos²x) + cos²x + 2cosx = 0
cos²x - 1 + cos²x + cos²x + 2cosx = 0
3cos²x + 2cosx - 1 = 0
Пусть t = cosx
3t² + 2t - 1 = 0
D = 16
t1 = 1/3
t2 = -1
cosx = 1/3
x = +-arccos(1/3) + 2πk
cosx = -1
x = π + 2πk