Determine the greatest possible value of \sum_{i=1}^{10} \cos 3x_i for real numbers PS

Determine the greatest possible value of
$\sum _{i=1}^{10}\mathrm{cos}3{x}_{i}$
for real numbers ${x}_{1},{x}_{2}\dots {x}_{10}$ satisfying
$\sum _{i=0}^{10}{\mathrm{cos}x}_{i}=0$
My attempt:
$\sum \mathrm{cos}3x=\sum 4{\mathrm{cos}}^{3}x-\sum 3\mathrm{cos}x=4\sum {\mathrm{cos}}^{3}x$
So now we have to maximize sum of cubes of ten numbers when their sum is zero and each lie in interval [−1,1]. i often use AM GM inequalities but here are 10 numbers and they are not even positive. Need help to how to visualize and approach these kinds of questions.
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Visualising the solution
You have asked for help in visualising the solution. I think you will find it useful to have in mind the picture of
Now consider the arrangement of the 10 numbers in the maximum position. (We have a continuous function on a compact set and so the maximum is attained.)
First suppose that there is a number, s, smaller in magnitude than the least negative number l. Increasing l whilst decreasing s by the same amount would increase the sum of cubes and therefore cannot occur.
So, all the negative numbers are equal, to l say, and all the positive numbers are greater than |l|.
Now suppose that a positive number was not 1. Then increasing it to 1 whilst reducing one of the ls would increase the sum of cubes and therefore cannot occur.
Hence we need only consider the case where we have m 1s and 10−m numbers equal to $-\frac{m}{10-m}$
Not exactly what you’re looking for?
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Hint: Use the fact that
$\mathrm{cos}a+\mathrm{cos}b=2\mathrm{cos}\left(\frac{12}{a+b}\right)\mathrm{cos}\left(\frac{12}{a-b}\right)$
If you pair the summands and apply above transformation, then the sum becomes a product with 10 cosine factors, and a scaling factor of ${2}^{5}$. So now at least we have an estimate of the sum