Question

2sin(3x-пи делённое на 4)=-корень из 2

Answer 1

$2 \sin(3x - \frac{\pi}{4} ) = - \sqrt{2} \\ \sin(3x - \frac{\pi}{4} ) = - \frac{ \sqrt{2} }{2} \\ 3x - \frac{\pi}{4} = ( - 1)^{n} \arcsin( - \frac{ \sqrt{2} }{2} ) + \pi n \\ 3x - \frac{\pi}{4} = ( - 1)^{n + 1} \frac{\pi}{4} + \pi n \\ 3x = ( - 1)^{n + 1} \frac{\pi}{4} + \frac{\pi}{4} + \pi n \\ x = ( - 1)^{n + 1} \frac{\pi}{12} + \frac{\pi}{12} + \frac{\pi}{3} n \\ x = ( {( - 1)}^{n + 1} + 1) \frac{\pi}{12} + \frac{\pi}{3} n, \: n \in Z$