Question

Решените уравнение
2sin(4x-п/6)+√2=0

Answer 1

$2 \sin(4x - \frac{\pi}{6} ) = - \sqrt{2} \\ \sin(4x - \frac{\pi}{6} ) = - \frac{ \sqrt{2} }{2}$

1 случай:

$4x - \frac{\pi}{6} = \frac{5\pi}{4} + 2\pi n \\ 4x = \frac{5\pi}{4} + \frac{\pi}{6} + 2\pi n \\ x = \frac{5\pi}{16} + \frac{\pi}{24} + \frac{\pi}{2} n \\ x = \frac{17\pi}{48} + \frac{\pi n}{2}, nєZ$

2 случай:

$4x - \frac{\pi}{6} = \frac{7\pi}{4} + 2\pi n \\ 4x = \frac{7\pi}{4} + \frac{\pi}{6} + 2\pi n \\ x = \frac{7\pi}{16} + \frac{\pi}{24} + \frac{\pi}{2} n \\ x = \frac{23\pi}{48} + \frac{\pi n}{2} , nєZ$