(One minor detail: you've written the theorem kind of backwards. Either it should be "linearly dependent", or your zero Wronskians should be nonzero.)
There are two different ways you could think about this.
First, any set of sufficiently differentiable functions whose Wronskian is nonzero at a point will be linearly independent; it's not necessary that those functions satisfy a nonsingular differential equation. (The hypothesis that they satisfy a DE is required for the "only if" half of the quoted theorem. For example, the set of functions $\{x^2, x^3\}$ has a Wronskian which vanishes at $0$, but is linearly independent.) Hopefully you should be able to look at the proof of quoted theorem in your textbook and convince yourself of this, or maybe it's even proved somewhere else in the book.
Second, you could also just apply the theorem as you've stated it; all you need to do is concoct a differential equation which has your functions as solutions! For example, if $D=\frac{d}{dx}$ is the differentiation operator, $1$ is a solution to $Dy=0$, and $\{\sin x, \cos x\}$ are solutions to $(D^2+1)y=0$, so $\{1, \sin x, \cos x\}$ must all be solutions to $D(D^2+1)y=0$. You can extend this idea to get an equation which works for $\sin 2x$ and $\cos 2x$ as well...