Question

Решите уравнение 1+cos2x+sin2x=0 и найдите корни из промежутка (-3П; -2П)

Answer 1

!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

там д.б cosx(tgx+1)=0 вместо 1+tgx=0

Answer 3

а)

$1+cos 2x+sin 2x=0medskip\sin^2x+cos^2x+cos^2x-sin^2x+2sin xcos x=0medskip\2cos^2x+2sin xcos x=0medskip\cos xleft(cos x+sin xright)=0medskip\1)~cos x=0medskip\x=dfrac{pi}{2}+pi k,~kinmathbb{Z}medskip\2)~cos x+sin x=0 middivcos xneq 0medskip\1+mathrm{tg}~x=0medskip\mathrm{tg}~x=-1medskip\x=-dfrac{pi}{4}+pi t,~tinmathbb{Z}$

б)

$left(-3pi;-2piright)medskip\1)~-3pi<dfrac{pi}{2}+pi m<-2pi;~minmathbb{Z}medskip\-3{,}5<m<-2{,}5;~minmathbb{Z}Rightarrow m=-3medskip\x_1=dfrac{pi}{2}-3pi=-dfrac{5pi}{2}medskip\2)~-3pi<-dfrac{pi}{4}+pi k<-2pi;~kinmathbb{Z}medskip\-2{,}75<k<-1{,}75;~kinmathbb{Z}Rightarrow k=-2medskip\x_2=-dfrac{pi}{4}-2pi=-dfrac{9pi}{4}$

Ответ. а) $x=dfrac{pi}{2}+pi m,~minmathbb{Z} vee x=-dfrac{pi}{4}+pi k,~kinmathbb{Z}$; б) $left{-dfrac{5pi}{2};-dfrac{9pi}{4}right}$