Решите уравнение: 4cos^2 (x-pi/6) - 3=0

Применим формулу разности квадратов:

4cos² (x - π/6) - 3 = 0;

(2cos (x - π/6) - √3) (2cos (x - π/6) + √3) = 0;

Произведение равно нулю если:

1) 2cos (x - π/6) - √3 = 0;

2cos (x - π/6) = √3;

cos (x - π/6) = √3/2;

Найдем значение аргумента:

x - π/6 = ± arccos (√3/2) + 2πn, n ∈ Z;

x - π/6 = ± π/6 + 2πn, n ∈ Z;

x = ± π/6 + π/6 + 2πn, n ∈ Z;

x1 = π/6 + π/6 + 2πn, n ∈ Z;

x1 = π/3 + 2πn, n ∈ Z;

x2 = - π/6 + π/6 + 2πn, n ∈ Z;

x2 = 2πn, n ∈ Z;

2) 2cos (x - π/6) + √3 = 0;

2cos (x - π/6) = - √3;

cos (x - π/6) = - √3/2;

Найдем значение аргумента:

x - π/6 = ± arccos ( - √3/2) + 2πn, n ∈ Z;

x - π/6 = π ± arccos (√3/2) + 2πn, n ∈ Z;

x - π/6 = π ± π/6 + 2πn, n ∈ Z;

x = π ± π/6 + π/6 + 2πn, n ∈ Z;

x3 = π + π/6 + π/6 + 2πn, n ∈ Z;

x3 = 4π/3 + 2πn, n ∈ Z;

x4 = π - π/6 + π/6 + 2πn, n ∈ Z;

x4 = π + 2πn, n ∈ Z;

Ответ: x1 = π/3 + 2πn, n ∈ Z, x2 = 2πn, n ∈ Z, x3 = 4π/3 + 2πn, n ∈ Z, x4 = π + 2πn, n ∈ Z;