Differential Equations- Wronskian Fails? I was doing a problem where the goal was to find whether two functions: $$f(x) = \sin(2x) ,~~~~~ \text{and}~~~~~~ g(x) = \cos(2x)$$ are linearly independent or not using the wronskian. 
The problem is simple enough, and after evaluating the wronskian I got a result of $~-2~$. Therefore, the implication is that the two functions are linearly independent for any value of $~x~$. And the way we've defined linear dependence is the following: 
If we can find some $~c_1~$ and $~c_2~$ constants such that $$c_1~f(x) + c_2~g(x) = 0~,$$ where $~c_1~$ and $~c_2~$ can't both be zero, then $~f(x)~$ and $~g(x)~$ are linearly dependent.
However, with this definition, I can easily think of a counter example to the result of our wronskian: If I pick $~c_1 = 1~$ and $~c_2 = -1~$, then at $~x = \frac{\pi}{8}~$, we get that $~\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} = 0~$. 
So clearly, $~f(x)~$ and $~g(x)~$ are linearly dependent at $~x = \frac{\pi}{8}~$ and, in fact, any $~x~$ such that $~x = \frac{\pi}{8} +2 ~\pi~n~$ 
This contradicts the result of our wronskian, which implies that $~f(x)~$ and $~g(x)~$ are linearly independent for any $~x~$.
Did I miss something? 
Maybe I'm interpreting some definition the wrong way. 

Without using Wronskian:
$$
a\sin(2x)+b\cos(2x)=\sqrt{a^2+b^2}\sin(2x+\delta)=0,\quad\forall x\in[x_1,x_2]
$$
then $\sqrt{a^2+b^2}=0$ since $\sin$ cannot be identical zero on an interval. It makes $a=b=0$, thus, independent.
