How do you calculate the position of the antinodes on a wave? if for example i have a waveform with the formula $y=\sin(1.2x)+\sin(1.8x)$. The first 6 antinodes on my graph come up at around:
$$
x=\begin{cases}
1.01\\
2.97\\
4.65\\
5.80\\
7.48\\
9.46\\
\end{cases}
$$
but I can't find a way to find the exact positions.
 A: So, looking at the points, seems you just need to find extrema of your function. In your particular case you can just find derivative
$$f(x)=\frac{\text{d}y(x)}{\text{d}x}$$ and numerically solve $$f(x)=0$$ starting near the points you're interested in.
Here's an example of a couple of extrema of your function found via W|A.
A: With a waveform of $y(x) = A_1 \sin( \kappa_1 x) + A_2 \sin(\kappa_2 x) $ the terms can be re-arranged into
$$ y(x) = (A_1+A_2) \cos  \left( \frac{\kappa_2-\kappa_1}{2} x \right) \sin  \left( \frac{\kappa_2+\kappa_1}{2} x \right) + (A_2-A_1) \sin \left( \frac{\kappa_2-\kappa_1}{2} x \right) \cos \left( \frac{\kappa_2+\kappa_1}{2} x \right) $$
and since in your case you have $A_1=A_2=1$ then
$$ y(x) = 2 \sin(1.5 x) \cos(0.3 x) $$
with the nodes $ x_i =\left( \frac{2 i}{\kappa_1 + \kappa_2}\right) \pi$ and $ x_j = \left( \frac{2 j- 1}{\kappa_2 - \kappa_1} \right) \pi$.
$$x_{\rm nodes} = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3},2\pi,\frac{8\pi}{3}, \frac{10\pi}{3} \ldots $$
Now you have the bounds for which you can find the anti-nodes numerically.
with the derivative $y'(x) = \frac{{\rm d}y(x)}{{\rm d}x} $
$$ y'(x) = \kappa_1 \cos(\kappa_1 x) + \kappa_2 \cos(\kappa_2 x) $$
you can do Newton-Raphson iteration starting from the mid-point between two nodes which should get you close to the anti-node. Otherwise you might end up near where $y''(x)=0$ causing an unstable solution.
