now, we can see that there is atleast one point in (1,2) where curve is cutting x axis..
this means (1+cos8x)(ax2+bx+c) = 0 for some x$\epsilon$(1,2)....., but,
(1+cos8x)$\neq$0 , since 1+cos8x>=1 $\forall$ x , therefore
(ax2+bx+c)=0 for some x$\epsilon$(1,2)...
hence (ax2+bx+c) has atleast one root in (0,2)..