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Katy256
@Katy256
July 2022
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Помогите, пожалуйста, решить уравнение:
(sinx-cosx)^2+tgx=2sin^2x
Заранее огромное спасибо!
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Answers & Comments
Lesben
(sinx-cosx)²+tgx=2sin²x
sin²x-2sinxcosx+cos²x+tgx =2sin²x
1-2.sinxcosx +sinx/cosx=2sin²x /cosx≠0
cosx-2sinxcos²x+sinx =2sin²xcosx
sinx+cosx=2sin²xcosx+2sinxcos²x
sinx+cosx=2sinxcosx(sinx+cosx) /:(sinx+cosx)≠0
1=2sinxcosx
sin2x=1
2x=π/2+2k.π
x=π/4+k.π , k∈Z
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Answers & Comments
sin²x-2sinxcosx+cos²x+tgx =2sin²x
1-2.sinxcosx +sinx/cosx=2sin²x /cosx≠0
cosx-2sinxcos²x+sinx =2sin²xcosx
sinx+cosx=2sin²xcosx+2sinxcos²x
sinx+cosx=2sinxcosx(sinx+cosx) /:(sinx+cosx)≠0
1=2sinxcosx
sin2x=1
2x=π/2+2k.π
x=π/4+k.π , k∈Z