Sinx=sqrt2/2 Тригонометрическое уравнение

$\sin x=\frac{\sqrt2}{2},\\\\x=(-1)^k\arcsin\frac{\sqrt2}{2}+\pi k,\ k\in Z,\\x=(-1)^k\bullet\frac{\pi}{4}+\pi k,\ k\in Z,\\\\1)\ k=2n,\\x=\frac{\pi}{4}+2\pi n,\ n\in Z;\\\\2)\ k=2n+1,\\x=-\frac{\pi}{4}+\pi(2n+1),\ n\in Z,\\x=-\frac{\pi}{4}+\pi+2\pi n,\ n\in Z,\\x=\frac{3\pi}{4}+2\pi n,\ n\in Z.$

Ответ: $x=\frac{\pi}{4}+2\pi n,\ x=\frac{3\pi}{4}+2\pi n,\ n\in Z.$