Solution:

By the product-to-sum identity, $2 \cos 4x \cos x = \cos 5x + \cos 3x,$ so the right side is $(\cos 5x + \cos 3x)^3.$ Setting $y = \cos 3x$ and $z = \cos 5x,$ the equation becomes $y^3 + z^3 = (y + z)^3,$ and since $(y+z)^3 - y^3 - z^3 = 3yz(y + z),$ it holds exactly when $$\cos 3x = 0, \qquad \cos 5x = 0, \qquad \text{or} \qquad \cos 4x \cos x = 0.$$

For $100 \lt x \lt 200$ in degrees: $\cos 3x = 0$ gives $x = 150;$ $\cos 5x = 0$ gives $x = 126,$ $162,$ $198;$ $\cos 4x = 0$ gives $x = 112.5,$ $157.5;$ and $\cos x = 0$ gives no solutions in the interval.

The sum is $150 + 126 + 162 + 198 + 112.5 + 157.5 = 906.$