What is the positive integer values of n such that `sinx=x/n` has 69 positive solutions?

Let's look at what the graphs of `sin(x)` and `x/n` might look like for a sample `n`, say `n=10`:

Notice that as long as `x/n<1`, two solutions (intersections) are generated on each interval `[2kpi, (2k+1)pi], k in ZZ^+`, that is, on each positive "hump" for `sin(x)`. As we are only looking for positive solutions, we can disregard `x=0`, meaning the first hump only gives a single solution.

With the above observations, we can tell that if the line `x/n` passes through `k` humps, then we generate `2k-1` positive solutions. As we are looking for `69` positive solutions, solving for `k` shows that we need `x/n` to pass through `35` humps.

Before continuing, note that `sin(x)=1` has no integral solutions, meaning `sin(x)=1` implies `x/n!=1`. This allows us to know that the line `x/n` either passes under `sin(x)` at the peak of a hump, thus generating two solutions, or passes over it, generating none.

If we look at the intervals on which `sin(x)` is positive, we find them to be of the form `[0, pi], [2pi, 3pi], [4pi, 5pi], ...`. The `35^"th"` interval will thus be `[68pi, 69pi]`. To get exactly `69` solutions, then, we must have `x/n < 1` at the midpoint of `[68pi, 69pi]` and `x/n > 1` at the midpoint of `[70pi, 71pi]`.

Put in terms of inequalities, this gives us

`((137pi)/2)/n < 1 < ((141pi)/2)/n`

`=> (137pi)/2 < n < (141pi)/2`

As `(137pi)/2 ~~ 215.2` and `(141pi)/2 ~~221.5`, we get our solution set as

`n in {216, 217, 218, 219, 220, 221}`