find the max+min

Let $f(x)$ be the given expression.

$\begin{aligned} f(x) & = 3\cos x -4\sin^2 x \cos x -2 \sin x +4 \sin 2x \cos x - 4 \sin 2x \sin x \\ & = 3\cos x -4\sin^2 x \cos x -2 \sin x + 8 \sin x \cos^2 x - 8 \sin^2 x \cos x \\ & = 3\cos x - 12\sin^2 x \cos x -2 \sin x + 8 \sin x \cos^2 x \\ & = 3\cos x - 12(1-\cos^2 x)\cos x -2 \sin x + 8(1-\sin^2 x)\sin x \\ & = -9\cos x + 12\cos^3 x + 6 \sin x - 8\sin^3 x \\ & = 3 \cos 3x + 2 \sin 3x \\ & = \sqrt{13} \sin \left(3x + \tan^{-1} \frac 32\right) \end{aligned}$

$\begin{aligned} \implies \max (f(x)) & = \sqrt{13} & \small \blue{\text{when }3x + \tan^{-1} \frac 32 = 2n\pi + \frac \pi 2} \\ \min (f(x)) & = - \sqrt{13} & \small \blue{\text{when }3x + \tan^{-1} \frac 32 = 2n\pi - \frac \pi 2} \\ \implies \max(f(x) + \min(f(x)) & = \sqrt{13} - \sqrt{13} = \boxed 0 & \small \blue{\text{where }n \text{ is an integer.}} \end{aligned}$

The given expression simplifies to $3\cos 3x-2\sin 3x=\sqrt {13}\cos (3x+\tan^{-1}(\frac{2}{3})$ .

So the minimum of the expression is $-\sqrt {13}$ , the maximum is $\sqrt {13}$ , and their sum is $\boxed 0$ .