1)3sinx*cosx+4cos^2x=0 2)3sin^2x+4cos^2x=13cosx*sinx

   1)

• 3sinx * cosx + 4cos^2x = 0;
• cosx(3sinx + 4cosx) = 0;

• [cosx = 0;
  [3sinx + 4cosx = 0;
• [cosx = 0;
  [tgx = -4/3;
• [x = /2 + k, k Z;
  [x = -arctg(4/3) + k, k Z.

   2)

• 3sin^2x + 4cos^2x = 13cosx * sinx;
• 3sin^2x - 13cosx * sinx + 4cos^2x = 0;
• 3tg^2x - 13tgx + 4 = 0;

• D = 13^2 - 4 * 3 * 4 = 169 - 48 = 121 = 11^2;
• tgx = (13 11)/6;

   a) tgx = (13 - 11)/6 = 2/6 = 1/3;

      x = arctg(1/3) + k, k Z.

   b) tgx = (13 + 11)/6 = 24/6 = 4;

      x = arctg(4) + k, k Z.

   Ответ:

• 1) /2 + k; -arctg(4/3) + k, k Z;
• 2) arctg(1/3) + k; arctg(4) + k, k Z.