Find the sum of the values of

Use the product to sum formula on the RHS,
$8\cos^34x\cos^3x=(2\cos4x\cos x)^3=(\cos5x+\cos3x)^3$
Factor the LHS,
$\cos^35x+\cos^33x=(\cos5x+\cos3x)(\cos^25x-\cos5x\cos3x+\cos^23x)$
So LHS=RHS if $\cos5x+\cos3x=2\cos4x\cos x=0$ which happens iff $x=22.5+45k,90,270$ where $0\leq k\leq7$ .
Now assume $\cos5x+\cos3x\neq0$ . Then we can divide it from both sides, and expand the RHS to get
$3\cos5x\cos3x=0$
Similarly, this has solutions $x=18+36k,30+60j$ where $0\leq k\leq9$ and $0\leq j\leq5$ .
Summing up all the solutions, we have
$\sum^7_{k=0}(22.5+45k)+\sum^9_{k=0}(18+36k)+\sum^5 _{k=0}(30+60k)-90-270=3960$