Задача №17545: Уравнение

Решите уравнение $cos 4 x + cos 2 x + ctg^{2} x = 0 .$

cos 4 x + cos 2 x + ctg^{2} x = 0 ⟺ 2 cos^{2} 2 x + cos 2 x + \frac{cos 2 x}{sin ^{2} x} = 0 ⟺

cos 2 x (2 - 4 sin^{2} x + 1 + \frac{1}{sin ^{2} x}) = 0 ⟺ \frac{cos 2 x}{sin ^{2} x} (4 sin^{4} x - 3 sin^{2} x - 1) = 0 ⟺

\frac{cos 2 x ( sin ^{2} x - 1 ) ( 4 sin ^{2} x + 1 )}{sin ^{2} x} = 0 ⟺ ⎩ ⎨ ⎧ [cos 2 x = 0 sin x = \pm 1 sin x \neq = 0 ⟺ x = \frac{π}{4} + \frac{k π}{2} x = \frac{π}{2} + k π, k \in Z

Решите уравнение $cos 4 x + cos 2 x + ctg^{2} x = 0 .$

cos 4 x + cos 2 x + ctg^{2} x = 0 ⟺ 2 cos^{2} 2 x + cos 2 x + \frac{cos 2 x}{sin ^{2} x} = 0 ⟺

cos 2 x (2 - 4 sin^{2} x + 1 + \frac{1}{sin ^{2} x}) = 0 ⟺ \frac{cos 2 x}{sin ^{2} x} (4 sin^{4} x - 3 sin^{2} x - 1) = 0 ⟺

\frac{cos 2 x ( sin ^{2} x - 1 ) ( 4 sin ^{2} x + 1 )}{sin ^{2} x} = 0 ⟺ ⎩ ⎨ ⎧ [cos 2 x = 0 sin x = \pm 1 sin x \neq = 0 ⟺ x = \frac{π}{4} + \frac{k π}{2} x = \frac{π}{2} + k π, k \in Z

Задача №17545