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Yaroslav2000000001
@Yaroslav2000000001
September 2021
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Решите 6 номер пожалуйста на листочке
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Jokius
А) 2sinx=
√3
sinx=√3/2
x1=arcsin√3/2+2πk
x1=π/3+2πk
x2=π-arcsin√3/2+2πk
x2=2π/3+2πk
б) cos3x=1/2
3x=
+-arccos1/2+2πk
3x=+-π/3+2πk
x=+-π/9+2πk/3
в) ctg(x-π/4)=√3
x-π/4=arcctg√3+πk
x-π/4=π/6+πk
x=5π/12+πk
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Yaroslav2000000001
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Answers & Comments
sinx=√3/2
x1=arcsin√3/2+2πk
x1=π/3+2πk
x2=π-arcsin√3/2+2πk
x2=2π/3+2πk
б) cos3x=1/2
3x=+-arccos1/2+2πk
3x=+-π/3+2πk
x=+-π/9+2πk/3
в) ctg(x-π/4)=√3
x-π/4=arcctg√3+πk
x-π/4=π/6+πk
x=5π/12+πk