9. (16 points) The real number $a$ makes the equation $4^{x}-4^{-x}=2 \cos a x$ have exactly 2015 roots. How many roots does the equation $4^{x}+4^{-x}=2 \cos a x+4$ have for this $a$?
9. The equation $4^{x}+4^{-x}=2 \cos a x+4$ is equivalent to

If the variable substitutions $x=2 y$ and $x=-2 z$ are applied to the two equations in the final system, they both become the first equation. Therefore, each of them has 2015 roots. However, these two equations cannot have a common root. In fact, if there were a common root $x=x_{0}$, then $4^{-\frac{x_{0}}{2}}-4^{\frac{x_{0}}{2}}=0$ and $\cos \frac{a x_{0}}{2}=0$ would both hold, but these two conditions cannot be satisfied simultaneously. Therefore, these two equations cannot have a common root. Thus, they have a total of $2 \times 2015=4030$ roots.