Fourier transform to the wave equation

Consider a solution to the wave equation $ psileft(x,tright) $, then using Fourier transform, we can represent:

$ psileft(x,tright)=left(frac{1}{2pi}right)^{2}int_{-infty}^{infty}int_{-infty}^{infty}widetilde{psi}left(k,omegaright)e^{ileft(kx+wtright)}dkdw $

Now if we’ll apply this form into the wave equation $ frac{partial^{2}psi}{partial x^{2}}-frac{1}{c^{2}}frac{partial^{2}psi}{partial t^{2}}=0 $

We’ll get:

$ int_{-infty}^{infty}int_{-infty}^{infty}widetilde{psi}left(k,omegaright)e^{ileft(kx+wtright)}left(-k^{2}+frac{omega^{2}}{c^{2}}right)dkdw=0 $

Now according to my book, this obligates the term $ left(-k^{2}+frac{omega^{2}}{c^{2}}right) $ to be $0$.

I really cant understand why. My knowledge in Fourier transform is very low, we’ve just learned maybe $ 2 $ hours just getting familier with the equations and applying it to some basic physics exercise. But I want to understand in a more profound way.

As I understand (with my basic knowledge of just year of math learnings), taking a fourier transform is equvivalent to representing a vector in a vector space using orthogonal basis. Which means $ e^{ileft(kx+wtright)} $ those are forming the orthogonal vector basis and the inner product is probably the integrals $ int_{-infty}^{infty}int_{-infty}^{infty} domega dk$.

So why does $ left(-k^{2}+frac{omega^{2}}{c^{2}}right) $ has to be $ 0 $ in order for the equation to make sense? as I see it maybe the term $ widetilde{psi}left(k,omegaright) $ can also cause everything to be $ 0 $.

If someone can explain in a simple way which dosent require profound understanding of the math behind the scenes, I’ll be greatful. Thanks