How do you like your 2017, oscillating or logarithmic?

Since $-1 \leq \cos^{2017}(x) \leq 1$ , the roots of f(x) exists at $-1 \leq \log_{2017}(x) \leq 1$ $\frac{1}{2017} \leq x \leq 2017$ .

Divide the interval into subintervals $((k - 1)\pi, k\pi)$ for $k = 1, 2, ..., \lfloor \frac{2017}{\pi} \rfloor = 642$ .

(Note: It's fine to start with 0 instead of $\frac{1}{2017}$ since there is no root at $x \leq \frac{1}{2017}$ )

By Intermediate Value Theorem, there will be exactly one root inside each of these subintervals.

For the leftover interval $(642\pi, 2017)$ , since $\cos(642\pi) = 1$ , the trigonometric part of the function will be decreasing while the logarithmic part is still increasing. These two parts will intersect again one last time; therefore f(x) will have a root at this interval.

642 roots from the subintervals + 1 additional root from the leftover interval = $\boxed{643}$ total roots