Question

Решите неравенство sinx+cosx > 1

Answer 1

$\sqrt{2}\sin \left(\frac{\pi }{4}+x\right)>1\\\sin \left(\frac{\pi }{4}+x\right)>\frac{\sqrt{2}}{2}\\\arcsin \left(\frac{\sqrt{2}}{2}\right)+2\pi n<\left(\frac{\pi }{4}+x\right)<\pi -\arcsin \left(\frac{\sqrt{2}}{2}\right)+2\pi n\\\arcsin \left(\frac{\sqrt{2}}{2}\right)+2\pi n<\frac{\pi }{4}+x\quad \mathrm{\&}\quad \frac{\pi }{4}+x<\pi -\arcsin \left(\frac{\sqrt{2}}{2}\right)+2\pi n\\\\\frac{\pi }{4}+x>\arcsin \left(\frac{\sqrt{2}}{2}\right)+2\pi n\\\frac{\pi }{4}+x>\frac{\pi }{4}+2\pi n\\x>2\pi n$

$\frac{\pi }{4}+x<\pi -\arcsin \left(\frac{\sqrt{2}}{2}\right)+2\pi n\\\frac{\pi }{4}+x<\pi -\frac{\pi }{4}+2\pi n\\\frac{\pi }{4}+x-\frac{\pi }{4}<\pi -\frac{\pi }{4}+2\pi n-\frac{\pi }{4}\\x<-\frac{\pi }{2}+\pi +2\pi n\\x<\frac{\pi }{2}+2\pi n$

$x>2\pi n\quad \mathrm{ \& }\quad \:x<\frac{\pi }{2}+2\pi n\\2\pi n<x<\frac{\pi }{2}+2\pi n\\Interval: x \in \left(2\pi n,\:\frac{\pi }{2}+2\pi n\right)$