When is $\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$? I'm trying to classify (at least as fully as possible) the values of $t$ in $(0,\pi/2)$ for which the following equation has a solution for some natural number, $n$.
$$\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$$
Does anyone know of any results that will help me?
 A: I don't know if there is any closed form for the solutions, but by replacing $\sin\big((n+2)t\big)$ with $\sin nt \cos 2t + \cos nt \sin 2t$ and rearranging, your problem is equivalent to finding the solutions of $$\tan nt = \frac{n \sin 2t}{n + 2 - n \cos 2t}.$$ For any $n > 0$, the RHS is bounded, while the LHS crosses from $-\infty$ to $\infty$ exactly $2n$ times over a $2 \pi$ interval. Some experimentation suggests that there are indeed exactly $2n$ solutions; possibly you could bound the slopes of the two expressions to make this rigorous. This would also let you find bounds for the locations of the solutions, and easily find them numerically through a bisection procedure.
A: EDIT: See J.M.'s comment to Rahul's answer: it's better to use the "other" Chebyshev polynomials.
Square your identity: $$(n+2)^2 \sin^2 nt  = n^2 \sin^2 (n+2)t.$$ Convert from sines to cosines: $$(n+2)^2 + n^2 \cos^2 (n+2)t = n^2 + (n+2)^2 \cos^2 nt.$$ Now let $x = \cos t$ to get $$(n+2)^2 + n^2 T_{n+2}(x)^2 = n^2 + (n+2)^2 T_n(x)^2,$$ where the $T_m$ are the Chebyshev polynomials. We get a polynomial equation of degree $2(n+2)$ in $x$ which you can solve to find all solutions (you will need to check that the signs of the sines match).
A: EDIT: This solution is wrong! The error is highlighted below (thanks for Zaricuse and Rahul).
There is no solution.
Given $t$, define a function $$f(n) = \frac{\sin(nt)}{n}.$$
Its derivative is $$f'(n) = \frac{\cos(nt)nt - \sin(nt)}{n^2}.$$
If $f(n) = f(n+2)$ then the derivative must have a zero somewhere in between, say $f'(m) = 0$. That implies $$\tan(mt) = mt.$$ However, for $x \in (0,\pi)$ we have $\tan(x) > x$ (consider the Taylor expansion of $\tan$, for example). The unfortunate error is that although $t < \pi$, $x = mt$ is not necessarily less than $\pi$.
