Jonathan T. answered • 10/05/23
10+ Years of Experience from Hundreds of Colleges and Universities!
It appears there might be some confusion in your question. I'll try to clarify and address it.
You mentioned the equation \(vx[1+\sin^2(2\pi/x)] = 2vx\), and you're wondering why the upper range for \(x\) is \(\leq 2\sqrt{x}\). However, the equation you provided seems to have a mistake, as the equation simplifies to \(1 + \sin^2(2\pi/x) = 2\). This equation is not directly related to the upper range of \(x\).
To determine the upper range for \(x\) (\(x \leq 2\sqrt{x}\)), we need to consider the context or the problem statement in which this inequality is given. The specific range for \(x\) would depend on the problem's conditions and constraints.
If you can provide more information or clarify the problem statement, I'd be happy to help you determine the appropriate upper range for \(x\) based on the given conditions.
