Question

алгебра, 20 баллов, прошу

Answer 1

$\tt\displaystyle sin\bigg(\frac{\pi}{18}-8x\bigg) = 1\\\\\\\frac{\pi}{18}-8x = \frac{\pi}{2} + 2\pi n, ~~n \in Z\\\\\\-8x = \frac{\pi}{2} -\frac{\pi}{18} + 2\pi n, ~~n \in Z\\\\\\-8x = \frac{8\pi}{18} + 2\pi n, ~~n \in Z\\\\\\x = -\frac{\pi}{18} - \frac{\pi n}{4}, ~~n \in Z$