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Автор - dinamn999

cos2x+sinквадратх =sinx

Ответ

Автор - jokeryl
ОДЗ sinx-cosx>0, потом обе части в квадрат. cos2x-sin4x=1-sin2x; cos2x-1=sin4x-sin2x; -2sin^2 x=2sinx*cos3x; sinx=0 или cos3x+sinx=0

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