Q:

Does this trig equation have no solution? $\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$

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 $x$.

A:

There are infinitely many solutions. In particular, as $\sqrt 2 \sin(\sqrt 2 x)$ oscillates between $-\sqrt 2$ and $\sqrt 2$, while the other term only oscillates between $-1$ and $1$, there is a solution at least once per period of $\sin (\sqrt 2 x)$ [although generically, more often than that].
You won't be able to easily write down all the solutions. One is $x = 0$. For any other one, you can numerically approximate the solutions very quickly.
For instance, as $\sin \pi = 0$ and $\sin 2\pi = 0$, while $\sin \sqrt 2 \pi < 0$ and $\sin \sqrt 2 (2 \pi) > 0$, the intermediate value theorem gives that there is another zero between $\pi$ and $2\pi$ (and which therefore can be quickly found using binary search).

A:

The arguments are different,so your method does not work. It is addition of sine waves of two different frequencies.There are no periodic solutions, but a set of infinite number of only real roots. It is surprising that there are no complex roots. Such equations arise in finding roots in certain eigenvalue problems.

Numerical iteration procedures like Newton/Raphson are helpful.
EDIT1:
If someone posts an animation of rotating force vectors it would be instructive and fun.
The bigger and faster vector overtakes the slower one infinitely many times. 
These roots are produced whenever y-axis projection is equal in magnitude and 
opposite in sign.

A:

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 $\sqrt{2}-1\leq |z|\leq \sqrt{2}+1$.

