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Kateuchiysya
@Kateuchiysya
June 2022
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В чем здесь ошибка? (фото)
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natali3221
√2\2sinx-√2\2cosx=1
sinπ\4sinx-cosπ\4cosx=1
-cos(π\4+x)=1
cos(π\4+x)=-1
π\4+x=π+2πn n∈Z
x=π-π\4+2πn n∈Z
x=(3π\4)+2πn n∈Z
0 votes
Thanks 1
Kateuchiysya
через косинус разве нельзя решить?
Kateuchiysya
все я поняла
Kateuchiysya
минус потеряла, да?
natali3221
да, перед косинусом
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Answers & Comments
sinπ\4sinx-cosπ\4cosx=1
-cos(π\4+x)=1
cos(π\4+x)=-1
π\4+x=π+2πn n∈Z
x=π-π\4+2πn n∈Z
x=(3π\4)+2πn n∈Z