Question

Помогите решить 2sin^2 x-7cosx-5=0

Answer 1

2(1 - Cos²x) - 7Cosx -5 = 0
2 - 2Cos²x  - 7Cosx - 5 = 0
2Cos²x + 7Cosx +3 = 0
Cosx = y
2y² + 7y +3 = 0
D = 49 - 24 = 25
y₁= (-7+5)/4 = -1/2                 у₂= ( -7 -5)/4 = -3
Cosx = -1/2                            Cosx = -3
x = +-2π/3 + 2π, k ∈Z                 ∅

Answer 2

2(1-cos²x)-7cosx-5=0
2-2cos²x-7cosx-5=0
-2cos²x-7cosx-3=0
2cos²x+7cosx+3=0
cosx=y
2y²+7y+3=0
D=49-4*3*2=25
y₁=(-7-5)/4=-3
y₂=(-7+5)/4=1/2
cosx=-3 не подходит
cosx=-1/2
x=(-1)ⁿ2π/3+πk,  k∈Z