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Решение: 1) cosx+sinx=cos2x/(1-sin2x ) (1-2sinx*cosx)(sinx+cosx)=(cos²x-sin²x) (1-2sinx*cosx)(sinx+cosx)-(cosx+sinx)(cosx-sinx)=0 (sinx+cosx)(1-2sinx*cosx-cosx+sinx)=0 sinx+cosx=0 tgx=-1 x1=-π/4+πn 1-2sinx*cosx-cosx+sinx=0 (1-cosx)-(sin2x-sinx)=0 2sin²(x/2)-2sin(x/2)*cos(3x/2)=0 2sin(x/2)(sin(x/2)-cos(3x/2))=0 sin(x/2)=0 x/2=πn x2=2πn sin(x/2)-cos(3x/2)=0 sin(x/2)-sin(π/2-3x/2)=0 2sin(-π/4+x)*cos(π/4-x/2)=0 sin(-π/4+x)=0 -π/4+x=πn x3=π/4+πn cos(π/4-x/2)=0 π/4-x/2=π/2+πn -x/2=π/2-π/4+πn -x/2=π/4+πn x4=-π/2+2πn ОДЗ: 1-sin2x≠0 sin2x≠1 2x≠π/2+2πn x≠π/4+π Итак ответ: x1=-π/4+πn x2=2πn x3=-π/2+2πn

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