We have the equation
\begin{equation}
2\delta\sin(x)-\sin((N+1)x)+\delta^2\sin((N-1)x) = 0,
\end{equation}
where $N$ is a positive integer greater than $1$ and $\delta\in[0,1]$.
Now, numerical studies indicate that the solutions to this equation are always real, but we have not been able to prove it. One idea was to use Rouche's Theorem, but that would require us to know about the number of real solutions in e.g. the interval $[0,\pi]$, and it seems hard to implement anyway since the equation has three terms. Another idea was to rewrite the equation in a convenient way, let $x = a+bi$ and see that the only way to have the right hand side be a real number, was to put $b=0$, but that just generated another equation that was hard to solve. Finally, we tried to argue in the following way: We know that for $\delta=0$, the equation has only real solutions, and we know the number of them. If we increase $\delta$ and show that the maxima of the curve $\sin((N+1)x)+\delta^2\sin((N-1)x)$ in the interval $[0,\pi]$ lie above the curve $2\delta\sin(x)$ for all $\delta$ (or precisely on the curve for $\delta=1$), we will not "lose" any real solutions. But analyzing the maxima was not so straightforward either.
Best Answer
As per @Lutz comment, trigonometric manipulations reduce the problem to proving that:
$\sin((N+1)x)=δ[\sin x\pm\sin(Nx)]=2δ\sin\left(\frac{N\pm1}2x\right)\cos\left(\frac{N\mp1}2x\right)$ has only real solutions.
Case 1: $\sin((N+1)x)=δ[\sin x+\sin(Nx)]$ or $\sin\left(\frac{N+1}2x\right)\cos\left(\frac{N+1}2x\right)=\delta\sin\left(\frac{N+1}2x\right)\cos\left(\frac{N-1}2x\right)$ and ignoring the roots of $\sin\left(\frac{N+1}2x\right)=0$ which are real, we need to prove that:
$\cos\left(\frac{N+1}2x\right)=\delta\cos\left(\frac{N-1}2x\right)$ has only real roots for any $0 \le \delta \le 1$
Rewriting the above as $e^{i(N+1)x}+1=\delta (e^{iNx}+e^{ix})$ and putting $z=e^{ix}$ (and using that $x \in \mathbb R$ iff $|e^{ix}|=1$) the problem reduces to showing that:
$(P_{\delta}(z))=P(z)=z^{N+1}-\delta(z^N+z)+1$ has roots only on the unit circle.
But clearly $P$ is $(N+1)$ self-inversive (which means $P^*(z)=z^{N+1}\overline{P(\frac{1}{\bar z})}=P(z)$ or equivalently $a_{N+1-k}=\bar a_k, P(z)=\sum_{k=0}^{N+1} a_kz^k$), so by general theory we have that $P$ has all roots on the unit circle iff $P'$ has all its roots inside the closed unit disc (we will give a short proof of this at the end)
A simple computation gives $P'(z)=(N+1)z^N-\delta N z^{N-1}-\delta$ and the triangle inequality immediately shows that $|z|>1$ (so $|z^N|>1, |z|^N>|z|^{N-1}$) implies $|P'(z)| >0$ since $|\delta| \le 1$ so we are done!
Case 2: $\sin((N+1)x)=δ[\sin x-\sin(Nx)]$
Now (ignoring the real roots of $\cos\left(\frac{N+1}2x\right)=0$), we need to prove:
$\sin\left(\frac{N+1}2x\right)=-\delta\sin\left(\frac{N-1}2x\right)$ has only real roots for any $0 \le \delta \le 1$.
With $z=e^{ix}$ this translates to $Q(z)=z^{N+1}+\delta(z^N-z)-1$ has roots only on the unit circle.
If $N$ even that $Q(-z)=P_{-\delta}(z)$ so we are done by case 1 (where we showed that we only need $|\delta| \le 1$).
If $N$ is odd, pick $\omega^{N+1}=-1$ a non trivial root of $-1$ and then:
$Q_1(z)=Q(\omega z)=-z^{N+1}+\delta(-\bar \omega z^N-\omega z)-1$ is self inversive.
It follows that the proof in case $1$ goes through using the triangle inequality to show that $|z|>1$ implies $|Q_1'(z)|>0$ so the roots of $Q_1$ hence of $Q$ are all on the unit circle and we are done!
Lemma: $P$ of degree $N$ self-inversive ($a_{N-k}=\bar a_k, P(z)=\sum_{k=0}^{N} a_kz^k$), then $P$ has all the roots on the unit circle iff $P'$ has all the roots inside the closed unit disc (note that by definition $P(0) \ne 0$ since $P$ has actual degree $N$)
Let $R(z)=\frac{zP'(z)}{NP(z)-zP'(z)}=\frac {zP'(z)}{(zP'(z)^*)} $ (where as before $Q^*(z)=z^{N}\overline{Q(\frac{1}{\bar z})}$ for a degree $N$ polynomial, while the above equality follows from the fact that $P$ is self inversive). In particular if $z_k, k=1,..m-1, m \le N$ are the roots of $P'/P$ one gets by a simple computation that $R(z)=cz\Pi_{k=1}^{m-1}\frac{z-z_k}{1-\bar z_k z}, |c|=1$ so if $P'$ (hence $P'/P$ too) has roots only inside the closed unit disc (actually the above compuation shows that the roots of $P'$ on the unit circle are roots of $P$ too so they are not among the $z_k$), we have that $R$ is a Blaschke product with $R(0)=0$ so in particular a unit disc self-map, hence $R(z) \ne -1, |z|<1$.
But $\frac{zP'(z)}{P(z)}=\frac{NR(z)}{1+R(z)}$ so $R(z) \ne -1, |z|<1$ implies $P$ cannot have roots inside the open unit disc. As $P$ self inversive means that roots come in pairs $z, 1/\bar z$, it follows that $P$ can't have roots outside the unit circle either, and we are done with the lemma as the converse is trivial from the above computations.