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.