Решить уравнение cos12x=cos6x+sin6x

   Воспользуемся формулой двойного угла косинус:

      cos(2a) = cos^2(a) - sin^2(a);

• cos(12x) = cos(6x) + sin(6x);
• cos^2(6x) - sin^2(6x) - (cos(6x) + sin(6x)) = 0;
• (cos(6x) + sin(6x))(cos(6x) - sin(6x)) - (cos(6x) + sin(6x)) = 0;
• (cos(6x) + sin(6x))(cos(6x) - sin(6x) - 1) = 0;

      [cos(6x) + sin(6x) = 0;      [cos(6x) - sin(6x) - 1 = 0.

   a) cos(6x) + sin(6x) = 0;

• sin(6x) = -cos(6x);
• tg(6x) = -1;
• 6x = -π/4 + πk, k ∈ Z;
• x = -π/24 + πk/6, k ∈ Z.

   b) cos(6x) - sin(6x) - 1 = 0;

• cos(6x) - sin(6x) = 1;
• √2/2 * cos(6x) - √2/2 * sin(6x) = √2/2;
• cos(π/4) * cos(6x) - sin(π/4) * sin(6x) = √2/2;
• cos(6x + π/4) = √2/2;
• 6x + π/4 = ±π/4 + 2πk, k ∈ Z;
• 6x = -π/4 ± π/4 + 2πk, k ∈ Z;

• [6x = -π/4 - π/4 + 2πk, k ∈ Z;[6x = -π/4 + π/4 + 2πk, k ∈ Z;
• [6x = -π/2 + 2πk, k ∈ Z;[6x = 2πk, k ∈ Z;
• [x = -π/12 + πk/3, k ∈ Z;[x = πk/3, k ∈ Z.

   Ответ: -π/24 + πk/6; -π/12 + πk/3; πk/3, k ∈ Z.