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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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asnaseerBest ResponseYou've already chosen the best response.0
From inspection (\(x=0\)) (and therefore \(x=2n\pi\), \(n=0,1,2,...\)) would certainly be a solution to this, but can you confirm that this is not valid since you said "least POSITIVE solution" and I understand that to mean \(x\gt 0\)?
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.1
Then other than 0, are there any other possible pairs?
 2 years ago

malevolence19Best ResponseYou've already chosen the best response.0
Well we know for cos(2ax) to be 1 we need that: \[(2m)\frac{a \pi}{b}=2n \pi; n,m \in \mathbb{Z}\] \[a=\frac{bn}{m}; n,m \in \mathbb{Z}\] I'm not sure if you might be looking for a different form.
 2 years ago
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