Question

Sinx-cosx>1. Очень надо ПЖ ПОМОГИТЕ ТРИГОНОМЕТРИЯ

Answer 1

$Sinx-Cosx>1\\\\\frac{\sqrt{2} }{2}Sinx-\frac{\sqrt{2}}{2}Cosx>\frac{\sqrt{2}}{2}\\\\Cos\frac{\pi }{4}Sinx-Sin\frac{\pi }{4}Cosx>\frac{\sqrt{2}}{2}\\\\Sin(x-\frac{\pi }{4})>\frac{\sqrt{2}}{2}\\\\\frac{\pi }{4}+2\pi n<x-\frac{\pi }{4}<\frac{3\pi }{4}+2\pi n,n\in Z\\\\\frac{\pi }{4}+\frac{\pi }{4}+2\pi n<x<\frac{3\pi }{4}+\frac{\pi }{4}+2\pi n,n\in Z\\\\\frac{\pi }{2}+2\pi n<x<\pi +2\pi n,n\in Z$

$Otvet:\boxed{(\frac{\pi }{2}+2\pi n;\pi + 2\pi n),n\in Z}$