Контрольные задания > 1)3sinx+1=0 2)2 cos4x=1 3)√2sin(2x-π/4)=1 4) sin2/3x=1/2 5)cos(5x+π/3)=-√3/2
Вопрос:

1)3sinx+1=0 2)2 cos4x=1 3)√2sin(2x-π/4)=1 4) sin2/3x=1/2 5)cos(5x+π/3)=-√3/2

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Ответ:

Краткое пояснение: Необходимо решить тригонометрические уравнения, используя знания о значениях тригонометрических функций и методы решения уравнений.

І вариант

  1. 3sinx + 1 = 0

    3sinx = -1

    sinx = -1/3

    x = arcsin(-1/3) + 2πk, x = -arcsin(1/3) + 2πk, k ∈ Z

  2. 2cos4x = 1

    cos4x = 1/2

    4x = ±arccos(1/2) + 2πk, k ∈ Z

    4x = ±π/3 + 2πk, k ∈ Z

    x = ±π/12 + πk/2, k ∈ Z

  3. √2sin(2x - π/4) = 1

    sin(2x - π/4) = 1/√2

    2x - π/4 = arcsin(1/√2) + 2πk, 2x - π/4 = π - arcsin(1/√2) + 2πk, k ∈ Z

    2x - π/4 = π/4 + 2πk, 2x - π/4 = π - π/4 + 2πk, k ∈ Z

    2x = π/2 + 2πk, 2x = 3π/4 + 2πk, k ∈ Z

    x = π/4 + πk, x = 3π/8 + πk, k ∈ Z

  4. sin(2/3 * x) = 1/2

    2/3 * x = arcsin(1/2) + 2πk, 2/3 * x = π - arcsin(1/2) + 2πk, k ∈ Z

    2/3 * x = π/6 + 2πk, 2/3 * x = π - π/6 + 2πk, k ∈ Z

    2/3 * x = π/6 + 2πk, 2/3 * x = 5π/6 + 2πk, k ∈ Z

    x = 3/2 * (π/6 + 2πk), x = 3/2 * (5π/6 + 2πk), k ∈ Z

    x = π/4 + 3πk, x = 5π/4 + 3πk, k ∈ Z

  5. cos(5x + π/3) = -√3/2

    5x + π/3 = ±arccos(-√3/2) + 2πk, k ∈ Z

    5x + π/3 = ±5π/6 + 2πk, k ∈ Z

    5x = -π/3 ± 5π/6 + 2πk, k ∈ Z

    5x = -π/3 + 5π/6 + 2πk, 5x = -π/3 - 5π/6 + 2πk, k ∈ Z

    5x = 3π/6 + 2πk, 5x = -7π/6 + 2πk, k ∈ Z

    5x = π/2 + 2πk, 5x = -7π/6 + 2πk, k ∈ Z

    x = π/10 + 2πk/5, x = -7π/30 + 2πk/5, k ∈ Z

ІІ вариант

  1. 4sinx = 3

    sinx = 3/4

    x = arcsin(3/4) + 2πk, x = π - arcsin(3/4) + 2πk, k ∈ Z

  2. 2cos3x = √3

    cos3x = √3/2

    3x = ±arccos(√3/2) + 2πk, k ∈ Z

    3x = ±π/6 + 2πk, k ∈ Z

    x = ±π/18 + 2πk/3, k ∈ Z

  3. 2sin(3x - π/6) = -√3

    sin(3x - π/6) = -√3/2

    3x - π/6 = arcsin(-√3/2) + 2πk, 3x - π/6 = π - arcsin(-√3/2) + 2πk, k ∈ Z

    3x - π/6 = -π/3 + 2πk, 3x - π/6 = π + π/3 + 2πk, k ∈ Z

    3x = π/6 - π/3 + 2πk, 3x = π + π/3 + π/6 + 2πk, k ∈ Z

    3x = -π/6 + 2πk, 3x = 9π/6 + 2πk, k ∈ Z

    x = -π/18 + 2πk/3, x = 3π/6 + 2πk/3, k ∈ Z

  4. cos(6x + π/2) = 2

    Уравнение не имеет решений, так как значение косинуса всегда находится в диапазоне [-1, 1].

  5. sin(2π/6 - x) = 1/2

    sin(π/3 - x) = 1/2

    π/3 - x = arcsin(1/2) + 2πk, π/3 - x = π - arcsin(1/2) + 2πk, k ∈ Z

    π/3 - x = π/6 + 2πk, π/3 - x = π - π/6 + 2πk, k ∈ Z

    -x = -π/6 + 2πk, -x = 3π/6 + 2πk, k ∈ Z

    x = π/6 - 2πk, x = -π/2 - 2πk, k ∈ Z

Ответ: Выше приведены решения тригонометрических уравнений для обоих вариантов.

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