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Kayshka
@Kayshka
July 2022
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Уравнение: cos4x+sin2x=0 90
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oganesbagoyan
Verified answer
Уравнение:
cos4x+sin2x=0 ;
1-2sin²2x+sin2x=0 ;
2sin²2x
-sin2x -1=0 ; * * *
t =sin2x * * *
2t
² -t -1 =0 ;
t₁= -1/2 ;
t₂ =1.
а)
sin2x = -1/2 ⇒ 2x = (-1)^(n+1)*π/6+πn , n∈Z.
x = (-1)^(n+1)*π/12+πn/2 , n∈Z.
б)
sin2x =1 ⇒ 2x = π/2+2πn , n∈Z.
x = π/4+πn , n∈Z.
ответ: (-1)^(n+1)*π/12+πn/2 ; π/4+πn , n∈Z.
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Verified answer
Уравнение:cos4x+sin2x=0 ;
1-2sin²2x+sin2x=0 ;
2sin²2x-sin2x -1=0 ; * * * t =sin2x * * *
2t² -t -1 =0 ;
t₁= -1/2 ;
t₂ =1.
а)
sin2x = -1/2 ⇒ 2x = (-1)^(n+1)*π/6+πn , n∈Z.
x = (-1)^(n+1)*π/12+πn/2 , n∈Z.
б)
sin2x =1 ⇒ 2x = π/2+2πn , n∈Z.
x = π/4+πn , n∈Z.
ответ: (-1)^(n+1)*π/12+πn/2 ; π/4+πn , n∈Z.