Question

2 sin x = 1
х принадлежит [-3П;3П]
найти все корни и записать в пор. возрастания​

Answer 1

Ответ:

$x = \{{-\frac{11\pi}{6},\ -\frac{7\pi}{6},\ \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6},\ \frac{17\pi}{6} }\}$

Объяснение:

$2sin(x) = 1, \ x \in [-3\pi;\ 3\pi]\\\\2sin(x)=1\\\\sin(x)=\frac{1}{2}\\\\x=(-1)^narcsin(\frac{1}{2})+\pi n, n \in \mathbb{Z}\\\\x = (-1)^n\cdot\frac{\pi}{6} +\pi n, n \in \mathbb{Z}\\\\x \in [-3\pi;\ 3\pi]\\\\n = -4, \ x=(-1)^{-4}\cdot\frac{\pi}{6} -4\pi= \frac{\pi}{6} -4\pi = - \frac{23\pi}{6} \notin [-3\pi;\ 3\pi]\\\\n = -3, \ x=(-1)^{-3}\cdot\frac{\pi}{6} -3\pi= -\frac{\pi}{6} -3\pi = - \frac{19\pi}{6} \notin [-3\pi;\ 3\pi]$

$\\\\n = -2, \ x=(-1)^{-2}\cdot\frac{\pi}{6} -2\pi= \frac{\pi}{6} -2\pi = - \frac{11\pi}{6} \in [-3\pi;\ 3\pi]\\\\n = -1, \ x=(-1)^{-1}\cdot\frac{\pi}{6} -\pi= -\frac{\pi}{6} -\pi = - \frac{7\pi}{6} \in [-3\pi;\ 3\pi]\\\\n = 0, \ x=(-1)^{0}\cdot\frac{\pi}{6} = \frac{\pi}{6} \in [-3\pi;\ 3\pi]\\\\n = 1, \ x=(-1)^{1}\cdot\frac{\pi}{6} +\pi= -\frac{\pi}{6} +\pi = \frac{5\pi}{6} \in [-3\pi;\ 3\pi]$

$\\\\n = 2, \ x=(-1)^{2}\cdot\frac{\pi}{6} +2\pi= \frac{\pi}{6} +2\pi = \frac{13\pi}{6} \in [-3\pi;\ 3\pi]\\\\n = 3, \ x=(-1)^{3}\cdot\frac{\pi}{6} +3\pi= -\frac{\pi}{6} +3\pi = \frac{17\pi}{6} \in [-3\pi;\ 3\pi]\\\\n = 4, \ x=(-1)^{4}\cdot\frac{\pi}{6} +4\pi= \frac{\pi}{6} +4\pi = \frac{25\pi}{6} \notin [-3\pi;\ 3\pi]$

$x = \{{-\frac{11\pi}{6},\ -\frac{7\pi}{6},\ \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6},\ \frac{17\pi}{6} }\}$