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Змей24
@Змей24
August 2022
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[tex]log_{sin3x}(cosx-cos2x)=1[/tex] Задача повышенной сложности, 11 класс.
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au456
Verified answer
По свойству логарифма
cos(x)-cos(2x)=sin(3x)
ОДЗ sin(3x)>0 sin(3x)<>1
ОДЗ (0;π/6)U(π/6;π/3)U(2π/3;5π/6)U(5π/6;π)U(4π/3;3π/2)U(3π/2;5π/3)
2sin(3x/2)*sin(x/2)=2sin(3x/2)*cos(3x/2)
sin(3x/2)=0
3x/2=πn
x=2πn/3 Вне ОДЗ
sin(x/2)=cos(3x/2)
sin(x/2)-sin(π/2-3x/2)=0
2sin(x-π/4)*cos(π/4-x/2)=0
cos(π/4-x/2)=0
x=π/2+2πn Вне ОДЗ
sin(x-π/4)=0
x=π/4+πn 5π/4+2πn Вне ОДЗ
Ответ : x=π/4+2πn
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Answers & Comments
Verified answer
По свойству логарифмаcos(x)-cos(2x)=sin(3x)
ОДЗ sin(3x)>0 sin(3x)<>1
ОДЗ (0;π/6)U(π/6;π/3)U(2π/3;5π/6)U(5π/6;π)U(4π/3;3π/2)U(3π/2;5π/3)
2sin(3x/2)*sin(x/2)=2sin(3x/2)*cos(3x/2)
sin(3x/2)=0
3x/2=πn
x=2πn/3 Вне ОДЗ
sin(x/2)=cos(3x/2)
sin(x/2)-sin(π/2-3x/2)=0
2sin(x-π/4)*cos(π/4-x/2)=0
cos(π/4-x/2)=0
x=π/2+2πn Вне ОДЗ
sin(x-π/4)=0
x=π/4+πn 5π/4+2πn Вне ОДЗ
Ответ : x=π/4+2πn