6sinx+5cosx=6 нужно решение

   Воспользуемся формулами двойных углов синуса и косинуса:

• sin(2α) = 2sinα * cosα;
• cos(2α) = cos^2(α) - sin^2(α);

• 6sinx + 5cosx = 6;
• 5cosx - 6(1 - sinx) = 0;
• 5(cos^2(x/2) - sin^2(x/2)) - 6(cos^2(x/2) - 2cos(x/2)sin(x/2) + sin^2(x/2)) = 0;
• 5(cos(x/2) - sin(x/2))(cos(x/2) + sin(x/2)) - 6(cos(x/2) - sin(x/2))^2 = 0;
• (cos(x/2) - sin(x/2))(5cos(x/2) + 5sin(x/2) - 6cos(x/2) + 6sin(x/2)) = 0;
• (cos(x/2) - sin(x/2))(11sin(x/2) - cos(x/2)) = 0;

• [cos(x/2) - sin(x/2) = 0;[11sin(x/2) - cos(x/2) = 0;
• [sin(x/2) = cos(x/2);[11sin(x/2) = cos(x/2);
• [tg(x/2) = 1;[tg(x/2) = 1/11;
• [x/2 = π/4 + πk, k ∈ Z;[x/2 = arctg(1/11) + πk, k ∈ Z.
• [x = π/2 + 2πk, k ∈ Z;[x = 2arctg(1/11) + 2πk, k ∈ Z.

   Ответ: π/2 + 2πk; 2arctg(1/11) + 2πk, k ∈ Z.