Question

Решите неравенство sinx+cosx > 1

Answer 1

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

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

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