What Would be a Sinusoidal Function that can Only Have Roots at all the Points I Have Highlighted?

Question

I am interested in a function which can represent the roots of the two functions which they do not share in common. The red function is y=sin((pi/2)x) and the green one is y=sin((pi/3)x) If anyone could provide help that would be awesome. The sketch is the blue function, the shape isn't necessary, only the fact that is has the roots indicated. Sketch Sin(pi/2 x) and Sin(pi/3 x)

Answer 1

We can define a stepwise function as :

From x = 2 + 12$k$ to 4 + 12$k$ and $k\in\mathbb{Z}$: $$y = sin(2\pi(x-2)/2)$$ From x = 4 + 12$k$ to 8 + 12$k$ : $$y = sin(2\pi(x-4)/8)$$ From x = 8 + 12$k$ to 10 + 12$k$ : $$y = - sin(2\pi(x-8)/2)$$ From x = 10 + 12$k$ to 14 + 12$k$ : $$y = - sin(2\pi(x-10)/8)$$

For sure this is not the only one. We can adjust the slope to be equal at x = 2,4,8,10,... Also we can adjust amplitude.

Hope this simple one be of help.

EDIT

Here there is another one:

$$y = sin(2\pi(x-3)/12)+ 0,5 sin(2\pi(x-1)/4)$$

Answer 2

Here's a weird solution:

$$y = \frac{\sin(\frac\pi2 x)\sin(\frac\pi3 x)}{\sin^2(\frac\pi2 x) + \sin^2(\frac\pi3 x)}$$

We feed in $\sin(\frac\pi2 x)$ and $\sin(\frac\pi3 x)$ into the classic analysis counterexample $f(x,y) = \frac{xy}{x^2+y^2}$.

Answer 3

If the shape isn't necessary, then how about:

$$ f(x) = \begin{cases} 0 &\textrm{ if $x$ is a root of exactly one of the two functions}\\ 1 &\textrm{ otherwise} \end{cases} $$