$$2 \sin 2x + 2 \sin(-x) - 2 \cos(-x) + 1 = 0$$
$$2 \cdot 2 \sin x \cos x - 2 \sin x - 2 \cos x + 1 = 0$$
$$4 \sin x \cos x - 2 \sin x - 2 \cos x + 1 = 0$$
$$2 \sin x (2 \cos x - 1) - (2 \cos x - 1) = 0$$
$$(2 \sin x - 1) (2 \cos x - 1) = 0$$
$$2 \sin x - 1 = 0 \quad \text{или} \quad 2 \cos x - 1 = 0$$
$$\sin x = \frac{1}{2} \quad \text{или} \quad \cos x = \frac{1}{2}$$
Решения уравнения $$\sin x = \frac{1}{2}$$:
$$x = \frac{\pi}{6} + 2\pi n, \quad n \in \mathbb{Z}$$
$$x = \frac{5\pi}{6} + 2\pi k, \quad k \in \mathbb{Z}$$
Решения уравнения $$\cos x = \frac{1}{2}$$:
$$x = \frac{\pi}{3} + 2\pi m, \quad m \in \mathbb{Z}$$
$$x = -\frac{\pi}{3} + 2\pi l, \quad l \in \mathbb{Z}$$
$$x = \frac{\pi}{6} + 2\pi n$$
Пусть $$n = 2$$: $$x = \frac{\pi}{6} + 4\pi = \frac{25\pi}{6} \approx 13.09 \in \left[ \frac{5\pi}{2}; 4\pi \right]$$
$$x = \frac{5\pi}{6} + 2\pi k$$
Пусть $$k = 2$$: $$x = \frac{5\pi}{6} + 4\pi = \frac{29\pi}{6} \approx 15.18
otin \left[ \frac{5\pi}{2}; 4\pi \right]$$
$$x = \frac{\pi}{3} + 2\pi m$$
Пусть $$m = 2$$: $$x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3} \approx 13.61 \in \left[ \frac{5\pi}{2}; 4\pi \right]$$
$$x = -\frac{\pi}{3} + 2\pi l$$
Пусть $$l = 2$$: $$x = -\frac{\pi}{3} + 4\pi = \frac{11\pi}{3} \approx 11.52 \in \left[ \frac{5\pi}{2}; 4\pi \right]$$
Ответ: a) $$x = \frac{\pi}{6} + 2\pi n, x = \frac{5\pi}{6} + 2\pi k, x = \frac{\pi}{3} + 2\pi m, x = -\frac{\pi}{3} + 2\pi l$$, где $$n, k, m, l \in \mathbb{Z}$$; б) $$\frac{25\pi}{6}, \frac{13\pi}{3}, \frac{11\pi}{3}$$