12sin (2X) + sinX+cosX=6

1. Пусть:

sinx + cosx = t.

Тогда:

t^2 = (sinx + cosx) ^2 = sin^2x + 2sinx * cosx + cos^2x = 1 + sin2x.

Отсюда:

sin2x = t^2 - 1.

2. Подставим в исходное уравнение:

12sin2x + sinx + cosx = 6; 12 (t^2 - 1) + t = 6; 12t^2 - 12 + t - 6 = 0; 12t^2 + t - 18 = 0; D = 1^2 + 4 * 12 * 18 = 865. t = (-1 ± √865) / 24. t1 = (-1 - √865) / 24 ≈ - 1,27; t2 = (-1 + √865) / 24 ≈ 1,18.

3. Обратная замена:

sinx + cosx = t; √2/2 * sinx + √2/2 * cosx = t√2/2; sin (x + π/4) = t√2/2; sin (x + π/4) = t√2/2; [x + π/4 = arcsin (t√2/2) + 2πk, k ∈ Z;

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

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