Задача №17728: Уравнение

Решите уравнение $sin 3 x (cos x - cos 2 x) - cos 3 x (sin x - sin 2 x) = 6 cos x - 3 .$

sin 3 x (cos x - cos 2 x) - cos 3 x (sin x - sin 2 x) = 6 cos x - 3 ⟺

⟺ (sin 3 x cos x - cos 3 x sin x) - (sin 3 x cos 2 x - cos 3 x sin 2 x) = 6 cos x - 3 ⟺

⟺ sin 2 x - sin x = 6 cos x - 3 ⟺ (2 cos x - 1) (sin x - 3) = 0 ⟺ cos x = \frac{1}{2} ⟺

⟺ x = \pm \frac{π}{3} + 2 k π, k \in Z

Решите уравнение $sin 3 x (cos x - cos 2 x) - cos 3 x (sin x - sin 2 x) = 6 cos x - 3 .$

sin 3 x (cos x - cos 2 x) - cos 3 x (sin x - sin 2 x) = 6 cos x - 3 ⟺

⟺ (sin 3 x cos x - cos 3 x sin x) - (sin 3 x cos 2 x - cos 3 x sin 2 x) = 6 cos x - 3 ⟺

⟺ sin 2 x - sin x = 6 cos x - 3 ⟺ (2 cos x - 1) (sin x - 3) = 0 ⟺ cos x = \frac{1}{2} ⟺

⟺ x = \pm \frac{π}{3} + 2 k π, k \in Z

Задача №17728