Найдите корни уравнения $$cos(4x + \frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$, принадлежащие промежутку $$[-\pi; \pi]$$

Решим уравнение: $$cos(4x + \frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$ Общее решение уравнения cos(t) = a, где |a| ≤ 1, имеет вид: t = ±arccos(a) + 2πn, n ∈ Z В нашем случае t = 4x + $$\frac{\pi}{4}$$, a = -$$\frac{\sqrt{2}}{2}$$ arccos(-$$\frac{\sqrt{2}}{2}$$) = $$\frac{3\pi}{4}$$ Тогда: 4x + $$\frac{\pi}{4}$$ = ±$$\frac{3\pi}{4}$$ + 2πn, n ∈ Z Выразим x: 4x = -$$\frac{\pi}{4}$$ ±$$\frac{3\pi}{4}$$ + 2πn, n ∈ Z x = -$$\frac{\pi}{16}$$ ±$$\frac{3\pi}{16}$$ + $$\frac{\pi}{2}$$n, n ∈ Z Рассмотрим два случая: 1) x = -$$\frac{\pi}{16}$$ + $$\frac{3\pi}{16}$$ + $$\frac{\pi}{2}$$n = $$\frac{2\pi}{16}$$ + $$\frac{\pi}{2}$$n = $$\frac{\pi}{8}$$ + $$\frac{\pi}{2}$$n, n ∈ Z 2) x = -$$\frac{\pi}{16}$$ - $$\frac{3\pi}{16}$$ + $$\frac{\pi}{2}$$n = -$$\frac{4\pi}{16}$$ + $$\frac{\pi}{2}$$n = -$$\frac{\pi}{4}$$ + $$\frac{\pi}{2}$$n, n ∈ Z Теперь найдем корни, принадлежащие промежутку [-π; π]: 1) x = $$\frac{\pi}{8}$$ + $$\frac{\pi}{2}$$n n = -1: x = $$\frac{\pi}{8}$$ - $$\frac{\pi}{2}$$ = $$\frac{\pi - 4\pi}{8}$$ = -$$\frac{3\pi}{8}$$ n = 0: x = $$\frac{\pi}{8}$$ n = 1: x = $$\frac{\pi}{8}$$ + $$\frac{\pi}{2}$$ = $$\frac{\pi + 4\pi}{8}$$ = $$\frac{5\pi}{8}$$ 2) x = -$$\frac{\pi}{4}$$ + $$\frac{\pi}{2}$$n n = -1: x = -$$\frac{\pi}{4}$$ - $$\frac{\pi}{2}$$ = -$$\frac{\pi + 2\pi}{4}$$ = -$$\frac{3\pi}{4}$$ n = 0: x = -$$\frac{\pi}{4}$$ n = 1: x = -$$\frac{\pi}{4}$$ + $$\frac{\pi}{2}$$ = $$\frac{-\pi + 2\pi}{4}$$ = $$\frac{\pi}{4}$$ n = 2: x = -$$\frac{\pi}{4}$$ + π = $$\frac{-\pi + 4\pi}{4}$$ = $$\frac{3\pi}{4}$$ Ответ: -$$\frac{3\pi}{4}$$, -$$\frac{3\pi}{8}$$, -$$\frac{\pi}{4}$$, $$\frac{\pi}{8}$$, $$\frac{\pi}{4}$$, $$\frac{5\pi}{8}$$, $$\frac{3\pi}{4}$$