Упр.728 ГДЗ Алимов 10-11 класс (Алгебра)

1) sin2x > =-1/2;
2) sin3x < корень 3/2.

1. $$\sin 2x \ge -\frac12$$

   Решим неравенство для аргумента:

   $$-\frac{\pi}{6}+2\pi n \le 2x \le \frac{7\pi}{6}+2\pi n,\quad n\in\mathbb Z$$

   Делим на $2$:

   $$-\frac{\pi}{12}+\pi n \le x \le \frac{7\pi}{12}+\pi n$$

   Пересечение с отрезком $$\left[-\frac{3\pi}{2};\pi\right]$$ даёт:

   $$\left[-\frac{3\pi}{2};-\frac{17\pi}{12}\right]\cup\left[-\frac{13\pi}{12};-\frac{5\pi}{12}\right]\cup\left[-\frac{\pi}{12};\frac{7\pi}{12}\right]\cup\left[\frac{11\pi}{12};\pi\right]$$

2. $$\sin 3x < \frac{\sqrt3}{2}$$

   Решим неравенство для аргумента:

   $$-\pi-\frac{\pi}{3}+2\pi n < 3x < \frac{\pi}{3}+2\pi n,\quad n\in\mathbb Z$$

   То есть

   $$-\frac{4\pi}{3}+2\pi n < 3x < \frac{\pi}{3}+2\pi n$$

   Делим на $3$:

   $$-\frac{4\pi}{9}+\frac{2\pi n}{3} < x < \frac{\pi}{9}+\frac{2\pi n}{3}$$

   Пересечение с отрезком $$\left[-\frac{3\pi}{2};\pi\right]$$ даёт:

   $$\left[-\frac{3\pi}{2};-\frac{11\pi}{9}\right)\cup\left(-\frac{10\pi}{9};-\frac{5\pi}{9}\right)\cup\left(-\frac{4\pi}{9};\frac{\pi}{9}\right)\cup\left(\frac{2\pi}{9};\frac{7\pi}{9}\right)\cup\left(\frac{8\pi}{9};\pi\right]$$

Ответ

1) $$\left[-\frac{3\pi}{2};-\frac{17\pi}{12}\right]\cup\left[-\frac{13\pi}{12};-\frac{5\pi}{12}\right]\cup\left[-\frac{\pi}{12};\frac{7\pi}{12}\right]\cup\left[\frac{11\pi}{12};\pi\right]$$

2) $$\left[-\frac{3\pi}{2};-\frac{11\pi}{9}\right)\cup\left(-\frac{10\pi}{9};-\frac{5\pi}{9}\right)\cup\left(-\frac{4\pi}{9};\frac{\pi}{9}\right)\cup\left(\frac{2\pi}{9};\frac{7\pi}{9}\right)\cup\left(\frac{8\pi}{9};\pi\right]$$