29.8. Сколько корней уравнения sin 3x = √2 принадлежит промежутку 2 3π π 2'2 [ ] ?

29.8. Сколько корней уравнения $$sin 3x = \frac{\sqrt{2}}{2}$$ принадлежит промежутку $$[ -\frac{3\pi}{2}; \frac{\pi}{2} ]$$?

$$3x = (-1)^n \frac{\pi}{4} + \pi n, n \in Z$$

$$x = (-1)^n \frac{\pi}{12} + \frac{\pi n}{3}, n \in Z$$

При $$n = -5$$: $$x = -\frac{\pi}{12} - \frac{5\pi}{3} = -\frac{\pi + 20\pi}{12} = -\frac{21\pi}{12} = -\frac{7\pi}{4} < -\frac{3\pi}{2}$$

При $$n = -4$$: $$x = -\frac{\pi}{12} - \frac{4\pi}{3} = -\frac{\pi + 16\pi}{12} = -\frac{17\pi}{12} > -\frac{3\pi}{2}$$

При $$n = -3$$: $$x = -\frac{\pi}{12} - \pi = -\frac{\pi + 12\pi}{12} = -\frac{13\pi}{12}$$

При $$n = -2$$: $$x = -\frac{\pi}{12} - \frac{2\pi}{3} = -\frac{\pi + 8\pi}{12} = -\frac{9\pi}{12} = -\frac{3\pi}{4}$$

При $$n = -1$$: $$x = -\frac{\pi}{12} - \frac{\pi}{3} = -\frac{\pi + 4\pi}{12} = -\frac{5\pi}{12}$$

При $$n = 0$$: $$x = \frac{\pi}{12}$$

При $$n = 1$$: $$x = \frac{\pi}{12} + \frac{\pi}{3} = \frac{\pi + 4\pi}{12} = \frac{5\pi}{12}$$

При $$n = 2$$: $$x = \frac{\pi}{12} + \frac{2\pi}{3} = \frac{\pi + 8\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4} > \frac{\pi}{2}$$

При $$n = 3$$: $$x = \frac{\pi}{12} + \pi = \frac{13\pi}{12} > \frac{\pi}{2}$$

При $$n = 4$$: $$x = \frac{\pi}{12} + \frac{4\pi}{3} = \frac{\pi + 16\pi}{12} = \frac{17\pi}{12} > \frac{\pi}{2}$$

Ответ: $$x_1 = -\frac{17\pi}{12}; x_2 = -\frac{13\pi}{12}; x_3 = -\frac{3\pi}{4}; x_4 = -\frac{5\pi}{12}; x_5 = \frac{\pi}{12}; x_6 = \frac{5\pi}{12}$$ - всего 6 корней

Ответ: 6