Confusion over $\not\equiv$ and $\neq$ when applying boundary conditions 
Here $k, A_1, A_2$ are constants
Although $A_1$cos$kx+A_2$sin$kx \not\equiv 0$, $A_1$cos$kx+A_2$sin$kx$ can equal zero at certain values of $x$ 
For example if $A_1=1, A_2=1, k=1$ then
cos$x$+sin$x=0$ at $x=-\pi/4$, so is the conclusion of this solution incorrect? i.e
Is the solution:
$w^2=gk$tanh$(kh)$ $\cup$ $A_1$cos$kx+A_2$sin$kx=0$?

The equation needs to hold for every $x$. For every choice of $A_1, A_2$ not both vanishing, there exists some $x$ such that $A_1 \cos kx + A_2 \sin kx \neq 0$. Therefore the other factor must be zero there. 
Since the other factor is independent of $x$, that's all we get, and all we need. 
