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: $$(2\cos(8x)−1)(2\cos(4x)−1)(2\cos(2x)−1)(2\cos(x)−1)=1 $$

Question

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: $$(2\cos(8x)−1)(2\cos(4x)−1)(2\cos(2x)−1)(2\cos(x)−1)=1 $$

asnaseer · Answer

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 $x\gt 0$?

anonymous · Answer

Then other than 0, are there any other possible pairs?

anonymous · Answer

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.