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moneybird

  • 2 years ago

What is the ordered pair of positive intgers (a,b) for which a/b is a reduced fraction and \[x = \frac{a \pi}{b}\] is the least positive solution of the equation: \[(2 \cos (8x) - 1)(2 \cos (4x) - 1)(2 \cos (2x) - 1)(2 \cos (x) - 1) = 1\]?

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  1. agdgdgdgwngo
    • 2 years ago
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    0?

  2. moneybird
    • 2 years ago
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    i don't know the solution

  3. agdgdgdgwngo
    • 2 years ago
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    \( 2 \pi \) ?

  4. moneybird
    • 2 years ago
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    how do you get that?

  5. Mertsj
    • 2 years ago
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    What course are you taking moneybird that this problem came from?

  6. moneybird
    • 2 years ago
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    Grade 10 math

  7. Mertsj
    • 2 years ago
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    ty

  8. Mertsj
    • 2 years ago
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    This equation will be true if each factor = 1. Each factor will equal 1 if x = 0. So a/b, which must be 2 positive integers is 2/1

  9. moneybird
    • 2 years ago
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    how about 1 factor is 4, and 1 factor is 1/4

  10. Mertsj
    • 2 years ago
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    That could not be true. The largest value of the cos is 1

  11. Mertsj
    • 2 years ago
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    So the largest value of 2cos x is 2

  12. jadest
    • 2 years ago
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    The maximum of every parenthetical is 1, the minimum is -2. You would need multiplicative inverses and even factors of positive for this to work

  13. Mertsj
    • 2 years ago
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    Regardless of the value of x

  14. moneybird
    • 2 years ago
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    Let me think about it

  15. sunsetlove
    • 2 years ago
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    good work

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