Question

$\sqrt{2} sin(2x - \frac{n}{6} ) = 1$
пожалуйста мне срочно надо​

Answer 1

$\sqrt{2}Sin(2x-\frac{\pi }{6})=1\\\\Sin(2x-\frac{\pi }{6})=\frac{1}{\sqrt{2}}\\\\2x-\frac{\pi }{6}=(-1)^{n}arc Sin\frac{1}{\sqrt{2}}+\pi n,n\in Z\\\\2x-\frac{\pi }{6}=(-1)^{n}\frac{\pi }{4} +\pi n,n\in Z\\\\2x=(-1)^{n}\frac{\pi }{4}+\frac{\pi }{6}+\pi n,n\in Z\\\\\boxed{x=(-1)^{n}\frac{\pi }{8}+\frac{\pi }{12}+\frac{\pi n }{2} ,n\in Z}$