Question

Решите уравнение 2sin^2х+7cosх+2=0

Answer 1

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