2sin2xcos2x-sin2x=0
sin2x=0
x=Пk/2
cos2x=1/2
x=+-П/6+Пk
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
2sin2xcos2x-sin2x=0
sin2x=0
x=Пk/2
cos2x=1/2
x=+-П/6+Пk
2(sinxcosx) ( cos^2к - sin^2х ) + sin^2x