2cos^2x/2-3 sin(п-x/2)cos(2п-x/2)+7sin^2x/2=3

   1. Преобразуем уравнение по формулам приведения:

• 2cos^2(x/2) - 3sin(π - x/2)cos(2π - x/2) + 7sin^2(x/2) = 3;
• 2cos^2(x/2) - 3sin(x/2)cos(x/2) + 7sin^2(x/2) = 3sin^2(x/2) + 3cos^2(x/2);
• 4sin^2(x/2) - 3sin(x/2)cos(x/2) - cos^2(x/2) = 0.

   2. Разделим обе части на cos^2(x/2):

      4tg^2(x/2) - 3tg(x/2) - 1 = 0;

      D = 3^2 + 4 * 4 = 25;

      tg(x/2) = (3 ± √25)/8 = (3 ± 5)/8;

   1) tg(x/2) = (3 - 5)/8 = -2/8 = -1/4;

• x/2 = -arctg(1/4) + πk, k ∈ Z;
• x = -2arctg(1/4) + 2πk, k ∈ Z;

   2) tg(x/2) = (3 + 5)/8 = 8/8 = 1;

• x/2 = π/4 + πk, k ∈ Z;
• x = π/2 + 2πk, k ∈ Z.

   Ответ: -2arctg(1/4) + 2πk; π/2 + 2πk, k ∈ Z.