6. $\frac{\sqrt{5}}{5}$.

According to the principal value of the inverse trigonometric function, we know $x>0$, the original equation is $\arcsin x-\arccos x=\arccos 2 x- \arcsin 2 x$. Taking the sine and rearranging, we get $x^{2}=\frac{7}{5}$. Since $x>0$, we have $x=\frac{\sqrt{5}}{5}$.
6. The solution to the equation $\arcsin x+\arcsin 2 x=\arccos x+\arccos 2 x$ is