пожалуйста помогите решить уравнение 3sin2х - 4cоsх + 3sinх - 2 = 0. Укажите корни ,принадледащие отрезку [ П/2 ; 3П/2 ]

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Lenok201

18 нояб. 2014 г., 21:18:31 (9 лет назад)

3sin2х - 4cоsх + 3sinх - 2 = 0

6sinх·cоsх - 4cоsх + 3sinх - 2 = 0

(6sinх·cоsх + 3sinх) - (4cоsх + 2) = 0

3sinх·(2cоsх + 1) - 2·(2cоsх + 1) = 0

(2cоsх + 1)·(3sinх - 2) = 0

1) 2cоsх + 1 = 0

cоsх = -1/2

x₁ = 4π/3 + 2πn

x₂ = -4π/3 + 2πn

2) 3sinх - 2 = 0

sinх = 2/3

x₃ = (-1)^k ·arcsin(2/3) + πk

Исследуем х₁ = 4π/3 + 2πn

n = 0 x₁ = 4π/3  x∈[π/2; 3π/2]

n = 1 x₁ = 4π/3 + 2π  x∉[π/2; 3π/2]

Исследуем x₂ = -4π/3 + 2πn

n = 1 x₂ = -4π/3 + 2π = 2π/3  x∈[π/2; 3π/2]

n = 2  x₂ = -4π/3 + 4π = 8π/3  x∉[π/2; 3π/2]

Исследуем x₃ = (-1)^k ·arcsin(2/3) + πk

arcsin(2/3) ≈ 42°

n = 1  x₃ = -arcsin(2/3) + π ≈ 138° x∈[π/2; 3π/2]

n = 2  x₃ = arcsin(2/3) + 2π ≈ 402° x∉[π/2; 3π/2]

Ответ: в интервале x∈[π/2; 3π/2] уравнеие имеет три корня

x₁ = 4π/3, x₂ = 2π/3, x₃ = -arcsin(2/3) + π