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 $x$.

$\sqrt{2} \sin(\sqrt{2}x)$ goes from $-1$ to $1$ on the interval $[(2n-1/4) \pi/\sqrt{2}, (2n+1/4)\pi/\sqrt{2}]$ and from $1$ to $-1$ on the interval $[(2n+3/4)\pi/\sqrt{2}, (2n+5/4)\pi/\sqrt{2}]$.  On each of these intervals
the derivative of $\sqrt{2}\sin(\sqrt{2}x)$ has absolute value at least $\sqrt{2}$.  Thus $\sqrt{2} \sin(\sqrt{2}x) + \sin(x)$ is monotone and has exactly one zero in each of these intervals.
