TITLE: Does this trig equation have no solution? $\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$
QUESTION [5 upvotes]: Solve the following trig. equation for $x$ 
  $$\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$$

My try:
I divided by $\sqrt{\left(\sqrt2\right)^2+1^2}=\sqrt 3$,
$$\frac{\sqrt 2}{\sqrt 3}\sin\left(\sqrt 2x\right)+\frac{1}{\sqrt 3}\sin (x)=0$$
$$\sqrt{\frac{2}{3}}\sin\left(\sqrt 2x\right)+\frac{1}{\sqrt 3}\sin (x)=0$$
I got stuck here, I have no clue to proceed to find the values of $x$.

REPLY [2 votes]: Very interesting question. Here it is a summary of this answer:


There are an infinite number of real roots;
All the roots of the given function are real;
The number of roots in the interval $[0,T]$ is extremely close to $T\pi^{-1}\sqrt{2}$.


1.) Given $f(x)=\sqrt{2}\sin(x\sqrt{2})+\sin(x)$, it is enough to prove that $f(x)$ is positive/negative at infinite points of $\pi\mathbb{Z}$ to get that $f(x)$ has an infinite number of zeroes, since it is a continuous function. So it is enough to study the behaviour of $f(k\pi)=\sqrt{2}\sin\left(\frac{2\pi k}{\sqrt{2}}\right)$ or the behaviour of 
$$ e_k = \exp\left(2\pi i\cdot\frac{k}{\sqrt{2}}\right). $$
Since $\sqrt{2}\not\in\mathbb{Q}$, $\{e_k\}_{k\in\mathbb{Z}}$ is dense in the unit circle (much more actually, it is a equidistributed sequence). The projection $z\to \text{Im}(z)$ (bringing $e_k$ to $\sqrt{1/2}\,f(k\pi)$) preserves density as any continuous map, hence it follows that $f(x)$ takes an infinite number of positive/negative values over $\pi\mathbb{Z}$, hence has an infinite number of real roots.
2.) Additionally, all the roots of $f(x)$ are real. This is a consequence of the Gauss-Lucas theorem, since the entire function $f(x)$ has an antiderivative
$$ F(x) = -\cos(x)-\cos(x\sqrt{2}) = -2\cos\left(\frac{1+\sqrt{2}}{2}x\right)\cos\left(\frac{1-\sqrt{2}}{2}x\right) $$
with only real roots, hence $f(x)$ cannot have any root in $\mathbb{C}\setminus\mathbb{R}$.
3.) Using a bit of topological degree theory, we have that the number of zeroes $N(T)$ of $f(x)$ in the interval $[0,T]$ is extremely well-approximated by the real part of
$$ \frac{1}{\pi}\int_{0}^{T}\frac{2e^{\sqrt{2}it}+e^{it}}{\sqrt{2} e^{\sqrt{2}it}+e^{it}}\,dt = \frac{T\sqrt{2}}{\pi}-\frac{\sqrt{2}-1}{\pi}\int_{0}^{T}\frac{dt}{1+\sqrt{2}e^{(\sqrt{2}-1)it}}$$
where the last integral is bounded. It follows that $N(T)\sim \frac{T\sqrt{2}}{\pi}$.
Here it is a plot of the curve $\gamma(t)=\sqrt{2}e^{\sqrt{2}it}+e^{it}$ for $t\in[0,200]$:
$\hspace1.5in$
by the triangle inequality, such a curve is contained in the annulus $\sqrt{2}-1\leq |z|\leq \sqrt{2}+1$.