5. Решите уравнения:
a) $$cos(\frac{x}{2} + \frac{\pi}{4}) + 1 = 0$$
$$cos(\frac{x}{2} + \frac{\pi}{4}) = -1$$
$$\frac{x}{2} + \frac{\pi}{4} = \pi + 2\pi n, n \in Z$$
$$\frac{x}{2} = \pi - \frac{\pi}{4} + 2\pi n, n \in Z$$
$$\frac{x}{2} = \frac{3\pi}{4} + 2\pi n, n \in Z$$
$$x = \frac{3\pi}{2} + 4\pi n, n \in Z$$
б) $$sin^2(x) - 2cos(x) + 2 = 0$$
$$1 - cos^2(x) - 2cos(x) + 2 = 0$$
$$cos^2(x) + 2cos(x) - 3 = 0$$
Пусть $$cos(x) = t$$, тогда:
$$t^2 + 2t - 3 = 0$$
$$D = 4 + 12 = 16$$
$$t_1 = \frac{-2 + 4}{2} = 1$$
$$t_2 = \frac{-2 - 4}{2} = -3$$
$$cos(x) = 1$$ или $$cos(x) = -3$$.
$$cos(x) = -3$$ - не имеет решений, так как $$-1 <= cos(x) <= 1$$.
$$cos(x) = 1$$
$$x = 2\pi n, n \in Z$$
Ответ: a) $$x = \frac{3\pi}{2} + 4\pi n, n \in Z$$; б) $$x = 2\pi n, n \in Z$$.