cos²(π-x)-sinx(π/2-x)=0
(-cosx)²-cosx=0
cos²x-cosx=0
cosx*(cosx-1)=0
cosx=0
x₁=π/2+πn.
cosx-1=0
cosx=1
x₂=2πn.
Ответ: x₁=π/2+πn x₂=2πn.
cos^2 (π-x) - sin (π/2-x)=0
cos(π-x)=-cosx
sin (π/2-x)=cosx
cos^2 (π-x) - sin (π/2-x)=(cosx)^2-cosx=0
cosx(cosx-1)=0
x1=pi/2+pi×k, k-целое
x2=2(pi)×k, k-целое
Ответ:
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
cos²(π-x)-sinx(π/2-x)=0
(-cosx)²-cosx=0
cos²x-cosx=0
cosx*(cosx-1)=0
cosx=0
x₁=π/2+πn.
cosx-1=0
cosx=1
x₂=2πn.
Ответ: x₁=π/2+πn x₂=2πn.
Verified answer
cos^2 (π-x) - sin (π/2-x)=0
cos(π-x)=-cosx
sin (π/2-x)=cosx
cos^2 (π-x) - sin (π/2-x)=(cosx)^2-cosx=0
cosx(cosx-1)=0
cosx=0
cosx-1=0
x1=pi/2+pi×k, k-целое
cosx=1
x2=2(pi)×k, k-целое
Ответ:
x1=pi/2+pi×k, k-целое
x2=2(pi)×k, k-целое