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SP1K3
@SP1K3
July 2022
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Решите уравнение [tex]9^{1-cos6x} =3^{ctg^{-1}3x}[/tex]
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Dимасuk
Verified answer
9^(1 - cos6x) = 3^((ctg^(-1)3x)
3^(2 - 2cos6x) = 3^(tg3x)
2 - 2cos6x = 3tg3x
2 - (2 - 4sin²3x) = 3tg3x
4sin²3x = 3tg3x
4sin²3x = 3sin3x/cos3x
4sin²3xcos3x = 3sin3x
4sin²3xcos3x - 3sin3x = 0
sin3x(4sin3xcos3x - 1) = 0
sin3x = 0 - нет корней, т.к. при sin3x = 0 уравнение не имеет смысла.
4sin3xcos3x - 1 = 0
2sin6x = 1
sin6x = 1/2
6x = (-1)ⁿπ/6 + πn, n ∈ Z
x = (-1)ⁿπ/36 + πn/6, n ∈ Z.
Ответ: x = (-1)ⁿπ/36 + πn/6, n ∈ Z.
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Answers & Comments
Verified answer
9^(1 - cos6x) = 3^((ctg^(-1)3x)3^(2 - 2cos6x) = 3^(tg3x)
2 - 2cos6x = 3tg3x
2 - (2 - 4sin²3x) = 3tg3x
4sin²3x = 3tg3x
4sin²3x = 3sin3x/cos3x
4sin²3xcos3x = 3sin3x
4sin²3xcos3x - 3sin3x = 0
sin3x(4sin3xcos3x - 1) = 0
sin3x = 0 - нет корней, т.к. при sin3x = 0 уравнение не имеет смысла.
4sin3xcos3x - 1 = 0
2sin6x = 1
sin6x = 1/2
6x = (-1)ⁿπ/6 + πn, n ∈ Z
x = (-1)ⁿπ/36 + πn/6, n ∈ Z.
Ответ: x = (-1)ⁿπ/36 + πn/6, n ∈ Z.