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Ameya
@Ameya
July 2022
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Помогите с уравнением !!!!! sin2x=2-sin^2x
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sedinalana
Verified answer
2sinxcosx-2sin²x-2cos²x+sin²x=0/cos²x
tg²x-2tgx+2=0
tgx=a
a²-2a+2=0
D=4-8=-4<0
нет решения
0 votes
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Ameya
но тут должно быть решение
oganesbagoyan
Verified answer
Sin2x =2 - sin²x ;
2sin2x =3+(
1 - 2sin²x
) ;
2sin2x - cos2x =3 ⇒ { sin2x =1 ; cos2x = -1 . ⇒ x ∈ ∅ .
* * * sin²2x +
cos
²
2x =1 * * *
0 votes
Thanks 0
oganesbagoyan
можно и без вычисления
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Answers & Comments
Verified answer
2sinxcosx-2sin²x-2cos²x+sin²x=0/cos²xtg²x-2tgx+2=0
tgx=a
a²-2a+2=0
D=4-8=-4<0
нет решения
Verified answer
Sin2x =2 - sin²x ;2sin2x =3+(1 - 2sin²x) ;
2sin2x - cos2x =3 ⇒ { sin2x =1 ; cos2x = -1 . ⇒ x ∈ ∅ .
* * * sin²2x + cos²2x =1 * * *