5 (sinx) 2+8 cosx+1=|cosx| + (cosx) 2

решите

5-5 сos²x+8cosx+1=/cosx/+cos²x

6cos²x+/cosx/-8cosx-6=0

1) cosx<0⇒x∈ (π/2+2πn; 3π/2+2πn)

6cos²x-9cosx-6=0

2cos²x-3cosx-2=0

cosx=a

2a²-3a-2=0

D=9+16=25

a1 = (3-5) / 4=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn

a2 = (3+5) / 4=2∉[-1; 1]-нет корней

2) cosx≥0⇒x∈[-π/2+πn; π/2+πn]

6cos²x-7cosx-6=0

cosx=a

6a²-7a-6=0

D=49+144=193

a1 = (7-√193) / 12⇒cosx = (7-√193) / 12<0-нет решения

a2 = (7+√193) / 12⇒cosx = (7+√193) / 12∉[-1; 1]-нет решения

Ответ x=+-2π/3+2πn