Решите уравнение cos2x+cosx=0

Cos2x+cosx=0

2cos²x-1+cosx=0

2cos²x+cosx-1=0

cosx=t

2t²+t-1=0

√D=√ (1+4*2) = 3

t1 = (-1-3) / (2*2) = - 1

cosx=-1

x1=π+2πk, k∈Z

t2 = (-1+3) / (2*2) = 1/2

cosx=1/2

x=π/3+2πk, k∈Z

x=-π/3+2πk, k∈Z

Ответ: x1=π+2πk

x2=π/3+2πk

x3 = - π/3+2πk