No.  If $x = 3^n \pi$ then $|f(x)| = f(\pi) \neq 0$ 
(numerically it's about $0.466$), and $3^n \pi$ can be arbitrarily large.

Such a construction fails for $\prod_{k=1}^\infty \cos(x/2^k)$
because if some $x/2^k$ is nearly $\pi$, 
or more generally some *odd* multiple of $\pi$, then
$x/2^{k+1}$ is nearly a half-integral multiple of $\pi$
and thus has a very small cosine.