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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: \[(2 \cos (8x)  1)(2 \cos (4x)  1)(2 \cos (2x)  1)(2 \cos (x)  1) = 1\]?
 2 years ago
 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: \[(2 \cos (8x)  1)(2 \cos (4x)  1)(2 \cos (2x)  1)(2 \cos (x)  1) = 1\]?
 2 years ago
 2 years ago

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moneybirdBest ResponseYou've already chosen the best response.3
i don't know the solution
 2 years ago

MertsjBest ResponseYou've already chosen the best response.0
What course are you taking moneybird that this problem came from?
 2 years ago

MertsjBest ResponseYou've already chosen the best response.0
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
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.3
how about 1 factor is 4, and 1 factor is 1/4
 2 years ago

MertsjBest ResponseYou've already chosen the best response.0
That could not be true. The largest value of the cos is 1
 2 years ago

MertsjBest ResponseYou've already chosen the best response.0
So the largest value of 2cos x is 2
 2 years ago

jadestBest ResponseYou've already chosen the best response.2
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
 2 years ago

MertsjBest ResponseYou've already chosen the best response.0
Regardless of the value of x
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.3
Let me think about it
 2 years ago
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