4sin2x-3 sin(2x-п/3)=5

Решим уравнение и найдем его корни:

4 * sin (2 * x) - 3 * sin(2 * x - pi/3) = 5; 

4 * sin (2 * x) - 3 * (sin (2 * x) * cos (pi/3) - cos (2 * x) * sin (pi/3)) = 5; 

4 * sin (2 * x) - 3 * (sin (2 * x) * 1/2 - cos (2 * x) * √3/2) - 5 = 0; 

4 * sin (2 * x) - 3/2 * sin (2 * x) + 3/2 * √3 * cos (2 * x) - 5 = 0; 

5/2 * sin (2 * x) + 3/2 * √3 * cos (2 * x) - 5 * sin^2 x - 5 * cos^2 x = 0; 

5 * sin x * cos x + 3/2 * √3 * cos^2 x - 3/2 * √3 * sin^2 x - 5 * sin^2 x - cos^2 x = 0; 

5 * tg x + 3/2 * √3  - 3/2 * √3 * tg^2 x - 5 * tg^2 x - 1 = 0;  

tg^2 x * (3/2 * √3 + 5) - 5 * tg x + (1 - 3/2 * √3) = 0;  

D = 25 - 4 * (3/2 * √3 + 5) *  (1 - 3/2 * √3) = 25 - 4 * (3/2 * √3 - 27/4 + 5 - 15/2 * √3) = 25 - 4 * (-27/4 + 5 - 6 * √3) = 72.8; 

1) tg x1 = (5 + 8.53)/5.1 = 2.6; 

x = arctg (2.6) + pi * n; 

2) tg x1 = (5 - 8.53)/5.1 = -0.7;  

x = -arctg (0.7) + pi * n, n принадлежит Z.