Every function of the form:
$$\psi(x, t) = A\cos[k(x\pm ct) + \phi]$$
is a solution of the wave equation $$\frac{\partial^2\psi}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}$$
The equation is linear, and this means that the general solution is in fact, any linear combination of the possible solutions, something that we can express as follows:
$$\psi(x, t) = \int_{-\infty}^{+\infty}dk\ [A_+(k)\cos(k(x-ct)) + A_-(k)\cos(k(x+ct)) + B_+(k)\sin(k(x-ct)) + B_-(k)\sin(k(x+ct))\ ]$$
The above integral is just the Fourier series for $\psi$, with the wave equation enforcing $k^2 = \frac{\omega^2}{c^2}$.
Now, solutions of this type are often called wave packets, as by suitable choice of the coefficients, the $A$s and the $B$s, you can represent any pulse that travels according to the wave equation. These (pulse-like) wave packets propagate like disturbances with speed $c$.
To see this easily, let $A_-$, and $B_+$ and $B_-$ all vanish for some pulse $\psi_1(x, 0)$, so that we only have the first cosine terms. Let's assume that at $t=0$, the maximum of the pulse is at $x=0$, where $x-ct = 0$. This means that the maximum of the pulse occurs when the argument of all the cosines vanish i.e. when $x = ct$ (because the only position dependence is in the cosines). This means that the peak travels with speed $c$, and the same argument carries over (for this wave equation) to any other part with a particular value of $\psi$.
Here's a sample wave packet propagating according to the wave equation (taken from the linked article)

It's worth noting that this pulse's motion ultimately comes from the phase velocity (i.e. the change of phase $x-ct$) of each of the $\cos$ terms, which is the motion of those solutions themselves. They do, of course, travel locally - the regions of constant phase move with speed $c$; it is simply because of the fact that the waveform repeats periodically that you can't think of it as a disturbance that originates somewhere.