1) ctg^4 * 2x-4ctg^2 * 2x+3=02) 4cos^2 (x-pi/6) - 3=0

1) Выполним замену ctg²2x = у:

ctg⁴2x - 4ctg²2x + 3 = 0;

у² - 4 у + 3 = 0;

Вычислим дискриминант:

D = b² - 4ac = ( - 4) ² - 4 * 1 * 3 = 16 - 12 = 4;

D › 0, значит:

у1 = ( - b - √D) / 2a = (4 - √4) / 2 * 1 = (4 - 2) / 2 = 2 / 2 = 1;

у2 = ( - b + √D) / 2a = (4 + √4) / 2 * 1 = (4 + 2) / 2 = 6 / 2 = 3;

Тогда, если у1 = 1, то:

ctg²2x = 1;

1) ctg2x = - 1;

2 х = arcctg ( - 1) + πn, n ∈ Z;

2 х = π - arcctg (1) + πn, n ∈ Z;

2 х = π - π / 4 + πn, n ∈ Z;

х1 = 3π / 8 + π/2 * n, n ∈ Z;

2) ctg2x = 1;

2 х = arcctg (1) + πn, n ∈ Z;

х2 = π / 8 + π/2 * n, n ∈ Z;

если у2 = 3, то:

ctg²2x = 3;

1) ctg2x = - √3;

2 х = arcctg ( - √3) + πn, n ∈ Z;

2 х = π - arcctg (√3) + πn, n ∈ Z;

2 х = π - π / 6 + πn, n ∈ Z;

х3 = 5π / 12 + π2 * n, n ∈ Z;

2) ctg2x = √3;

2 х = arcctg (√3) + πn, n ∈ Z;

х4 = π/12 + π/2 * n, n ∈ Z;

Ответ: х1 = 3π / 8 + π/2 * n, n ∈ Z, х2 = π / 8 + π/2 * n, n ∈ Z, х3 = 5π / 12 + π2 * n, n ∈ Z, х4 = π/12 + π/2 * n, n ∈ Z.

2) Применим формулу понижения степени:

4cos² (x - π/6) - 3 = 0;

cos² (x - π/6) = (1 + cos2 (x - π/6)) / 2 = (1 + cos (2x - π/3)) / 2;

Подставим полученные значения:

4 * (1 + cos (2x - π/3)) / 2 - 3 = 0;

2 * (1 + cos (2x - π/3)) - 3 = 0;

2 + 2cos (2x - π/3) - 3 = 0;

2cos (2x - π/3) - 1 = 0;

2cos (2x - π/3) = 1;

cos (2x - π/3) = 1/2;

Найдем значение аргумента:

2x - π/3 = ± arccos (1/2) + 2πn, n ∈ Z;

2x - π/3 = ± π/3 + 2πn, n ∈ Z;

2x = ± π/3 + π/3 + 2πn, n ∈ Z;

1) 2x = - π/3 + π/3 + 2πn, n ∈ Z;

2x = 2πn, n ∈ Z;

x1 = πn, n ∈ Z;

2) 2x = π/3 + π/3 + 2πn, n ∈ Z;

2x = 2π/3 + 2πn, n ∈ Z;

x2 = π/3 + πn, n ∈ Z;

Ответ: x1 = πn, n ∈ Z, x2 = π/3 + πn, n ∈ Z;