9. The sum of 1265 natural numbers is $2016+33$, and their product is $-\left(2016^{2}+33\right)$. Find all possible sets of such numbers. In your answer, indicate the sum of the largest and smallest numbers from all the found sets. If such numbers do not exist, then in your answer, indicate the number 0.

Hint: $2016^{2}+33=3 \cdot 41 \cdot 173 \cdot 191$.
Answer 574.

II-A. Solve the equation

$$
(1+\cos x+\cos 2 x+\cos 3 x+\cos 4 x)^{4}+(\sin x+\sin 2 x+\sin 3 x+\sin 4 x)^{4}=\frac{3}{4}+\frac{\cos 8 x}{4}
$$

Find the sum of all roots on the interval $A$, rounding to the nearest integer if necessary. If there are no roots or infinitely many roots on this interval, write the digit 0 in the answer.

Solution. 1st method of solving. The following is true:

$$
\begin{aligned}
& (1+\cos 4 x)+(\cos x+\cos 3 x)+\cos 2 x=\cos 2 x(2 \cos 2 x+2 \cos 2 x+1) \\
& \sin 2 x+\sin 4 x+(\sin x+\sin 3 x)=\sin 2 x(1+2 \cos 2 x+2 \cos x)
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
& (2 \cos 2 x+2 \cos 2 x+1)^{4}\left(\cos ^{4} 2 x+\sin ^{4} 2 x\right)=\frac{3}{4}+\frac{1-2 \sin ^{2} 4 x}{4} \Longleftrightarrow \\
& \left(4 \cos ^{2} x+2 \cos x-1\right)^{4}\left(\cos ^{4} 2 x+\sin ^{4} 2 x\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(4 \cos ^{2} x+2 \cos x-1\right)^{4}\left(1-\frac{\sin ^{2} 4 x}{2}\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(4 \cos ^{2} x+2 \cos x-1\right)^{4}=1 \Longleftrightarrow \\
& 4 \cos ^{2} x+2 \cos x-1= \pm 1
\end{aligned}
$$

Solving the quadratic equations (in terms of the variable $\cos x$) $4 \cos ^{2} x+2 \cos x-2=0$ and $4 \cos ^{2} x+2 \cos x=0$, we arrive at $\cos x=1 / 2 ;-1 ; 0 ;-1 / 2$.

Answer: $x= \pm \frac{\pi}{3}+2 \pi n, x= \pm \frac{2 \pi}{3}+2 \pi n, x=\frac{\pi}{2}+\pi n, x=\pi+2 \pi n, n \in \mathbb{Z}$.

2nd method of solving. It is easy to see that there are no solutions on the set $x=2 \pi n, n \in \mathbb{Z}$. Consider all other values $x \in \mathbb{R}$. The equation is equivalent to:

$$
\begin{aligned}
& \left(\frac{\sin \frac{5 x}{2} \cos 2 x}{\sin \frac{x}{2}}\right)^{4}+\left(\frac{\sin \frac{5 x}{2} \sin 2 x}{\sin \frac{x}{2}}\right)^{4}=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(\frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}}\right)^{4}\left(\cos ^{4} 2 x+\sin ^{4} 2 x\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(\frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}}\right)^{4}\left(1-\frac{\sin ^{2} 4 x}{2}\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(\frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}}\right)^{4}=1
\end{aligned}
$$

Solve the equation

$\sin \frac{5 x}{2}= \pm \sin \frac{x}{2} \Longleftrightarrow\left[\begin{array}{l}\frac{5 x}{2}= \pm \frac{x}{2}+2 \pi n, \\ \frac{5 x}{2}=\pi \pm \frac{x}{2}+2 \pi m,\end{array} \Longleftrightarrow\right.$

$$
\left[\begin{array} { l } 
{ 3 x = 2 \pi n , } \\
{ 2 x = 2 \pi n , } \\
{ 3 x = \pi + 2 \pi n , } \\
{ 2 x = \pi + 2 \pi n }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{2 \pi n}{3} \\
x=\pi n \\
x=\frac{\pi}{3}+\frac{2 \pi n}{3} \\
x=\frac{\pi}{2}+\pi n
\end{array}\right.\right.
$$

The original equation on the set $x \neq 2 \pi n, n \in \mathbb{Z}$ is equivalent to

$$
\left[\begin{array}{l}
x=\frac{2 \pi m}{3}, m \in \mathbb{Z}, m \neq 3 s, s \in \mathbb{Z} \\
x=\frac{\pi}{3}+\frac{2 \pi n}{3}, n \in \mathbb{Z} \\
x=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}
\end{array}\right.
$$

Answer: On the set $A=[2 \pi m ; 2 \pi m+\pi]$. Answer $\frac{5 \pi}{2}+8 \pi m, m \in \mathbb{Z}$.