Question

2sin^2x-3sinxcosx-5cos^2x>0 ,срочно​

Answer 1

Ответ:

$2sin^2x-3sinx\, cosx-5cos^2x>0\ \Big|:cos^2x>0\\\\2tg^2x-3tgx-5>0\ \ ,\ \ D=49\ \ ,\ \ tgx=-1\ \ \ ili\ \ \ tgx=2,5\\\\2(tgx+1)(tgx-2,5)>0\\\\znaki:\ \ +++(-1)---(2,5)+++\\\\tgx\in (-\infty ;-1\ )\cup (\ 2,5\ ;+\infty )\\\\a)\ \ tgx<-1\ \ ,\ \ -\dfrac{\pi}{2}+\pi n<x<-\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\b)\ \ tgx>2,5\ \ ,\ \ \ arctg2,5+\pi n<x<\dfrac{\pi}{2}+\pi n\ ,\ n\in Z$

$Otvet:\ x\in \Big( -\dfrac{\pi}{2}+\pi n\ ;\ -\dfrac{\pi}{4}+\pi n\ \Big)\cup \Big(\ arctg2,5+\pi n\ ;\ \dfrac{\pi}{2}+\pi n\ \Big)\ ,\ n\in Z\ .$