Решите уравнение 2sin^2 х+7cosх+2=0

2sin²x+7cosx+2=0

2 (1-cos²x) + 7cosx+2=0

2-2cos²x+7cosx+2=0

-2cos²x+7cosx+4=0

2cos²x-7cosx-4=0

cosx=a

2a²-7a-4=0

D=49+4*2*4 = 49+32 = 81

a₁=7+9/4 = 4

a₂ = 7-9/4 = - 1/2

cosx=4

х=решения нет

cosx=-1/2

x = + - arccos (-1/2) + 2πn

x = + - (π-arccos1/2) + 2πn

x = + - (π-π/3) + 2πn

x = + - 2π/3+2πn, n∈z