1) 1-sin^4x-5cos^4x=0 2) 4sin^3x-(sinx+cosx)=0 - вопрос №27870581145402430 от Alesha55551 21.01.2021 13:20

Question

Алгебра: 1) 1-sin^4x-5cos^4x=0 2) 4sin^3x-(sinx+cosx)=0

Alesha55551 · Answer

2) 4sin^3x-(sinx+cosx)=0
4sin^3 x - sinx - cosx = 0
(4sin^3 x - sinx) - cosx = 0
4sin x * (sin^2 x - 1) - cosx = 0
-4sin x * (1- sin^2 x) + cosx = 0
-4sin x * cos^2 x + cosx = 0
cos x * (1 - 4sin x*cos x) = 0
Или cos x = 0 =>
x= п/2+2пn
Или 1 - 4sin x*cos x = 0 =>
1 - 4*1/2*(sin 0 + sin 2x) = 0
1 - 2sin 2x = 0
sin 2x = 1/2
2x = п/6 =>
x = п/12+ 2пn