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Neonn7
@Neonn7
July 2022
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Пожалуйста помогите с решением неравенства:
[tex]sin6x-4sin2x\ \textless \ 0[/tex]
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sedinalana
Verified answer
Sin(2x+4x)-4sin2x<0
sin2x*cos4x+sin4x*cos2x-4sin2x<0
sin2x*cos4x+2sin2x*cos²2x-4sin2x<0
sin2x*(cos4x+2cos²2x-4)<0
sin2x*(1-2sin²2x+2-2sin²2x-4)<0
sin2x*(-1-4sin²2x)<0
sin2x*(1+4sin²x)>0
1+4sin²2x>0 при любом х⇒
sin2x>0
2πk<2x<π+2πk
πk<x<π/2+πk
x∈(πk;π/2+πk,k∈z)
2 votes
Thanks 2
Neonn7
большое спасибо!
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Answers & Comments
Verified answer
Sin(2x+4x)-4sin2x<0sin2x*cos4x+sin4x*cos2x-4sin2x<0
sin2x*cos4x+2sin2x*cos²2x-4sin2x<0
sin2x*(cos4x+2cos²2x-4)<0
sin2x*(1-2sin²2x+2-2sin²2x-4)<0
sin2x*(-1-4sin²2x)<0
sin2x*(1+4sin²x)>0
1+4sin²2x>0 при любом х⇒
sin2x>0
2πk<2x<π+2πk
πk<x<π/2+πk
x∈(πk;π/2+πk,k∈z)