А) Решите уравнение 2*sin (x) * sin (5*Pi/2+2x) - 4*cos^2 (Pi+x) = sin (x) - 3; б) Найдите корни, принадлежащие промежутку [3*Pi/2; 3*Pi]

2sin x * sin (5π/2 + 2x) - 4cos² (π + x) = sin x - 3;

2sin x * sin (2π + π/2 + 2x) - 4cos² (π + x) = sin x - 3;

2sin x * sin (π/2 + 2x) - 4cos² (π + x) = sin x - 3;

2sin x * cos 2x - 4cos² x = sin x - 3;

2sin x * cos 2x - 2 * (1 + cos² x) = sin x - 3;

2sin x * cos 2x - 2cos 2x - sin x = - 1;

2cos 2x * (sin x - 1) - sin x + 1 = 0;

(sin x - 1) * (2cos 2x - 1) = 0;

sin x - 1 = 0 или 2cos 2x - 1 = 0;

sin x - 1 = 0;

sin x = 1;

x = π/2 + 2πK, K ϵ Z;

1. K = 1, x₁ = 5π/2.

2cos 2x - 1 = 0;

cos 2x = 1/2;

2x = ± π/3 + 2πK, K ϵ Z;

x = ± π/6 + πK, K ϵ Z;

2. K = 1, x₂ = 13π/6, K = 1, x₃ = 11π/6, K = 1, x₄ = 17π/6.

Ответ: корни уравнения: x₁ = 5π/2, x₂ = 13π/6, x₃ = 11π/6, x₄ = 17π/6.