Would you square the vector?

We don't know the order of the vectors, so we have to take two cases:

Case 1

$\implies \displaystyle |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$

$\implies \displaystyle |\vec{a}+\vec{b}| = |\vec{a}+(-\vec{b})|$

$\implies \displaystyle |\sqrt{a^{2}+b^{2}+2ab \cos x}|=|\sqrt{a^{2}+b^{2}+2a(-b) \cos x}|$

Now it is pretty evident that if we take the second case, it will be nothing else than the shifting of the above equation's LHS to the RHS and vice versa. So we can continue since the second case leads to the same equation. We now square the sides.

$\implies \displaystyle 2ab \cos x=-2ab \cos x$

$\implies \displaystyle \cos x= -\cos x$

$\implies \displaystyle 2\cos x=0$

$\implies \displaystyle x = \arccos(0) = 90^\circ$ $\implies \displaystyle \boxed{x=90^\circ}$

$|\vec a+\vec b|=|\vec a-\vec b|$
$(\vec a+\vec b)\cdot(\vec a+\vec b)=(\vec a-\vec b)\cdot (\vec a-\vec b)$
Simplifying yields $\vec a \cdot \vec b=0$
$\therefore 90^\circ$