If f(x) = cos [e] x + cos [–e] x,

Given: f(x) = cos[e]x + cos[–e]x

$f(x)=\cos [e] x+\cos [-e] x$

$=\cos 2 x+\cos (-3 x) \quad(\because e=2.718$ approx $)$

$=\cos 2 x+\cos 3 x$

$\Rightarrow f(\pi)=\cos 2 \pi+\cos 3 \pi$

$=1-1$

$=0$

Hence, $f(\pi)=\underline{0}$