Problem 6. Find the greatest negative root of the equation

$$
\sin (x)+8 \cos (x)=4 \sin (8 x)+7 \cos (8 x)
$$
Answer: $\frac{\alpha+\beta-2 \pi}{9} \approx-0.6266$, where $\alpha=\operatorname{arctg} \frac{1}{8}, \beta=\operatorname{arctg} \frac{4}{7}$.

Comment 1. Depending on the method of solution chosen and the inverse trigonometric function used, the answer can take different forms. Let's present a few more ways to describe $\alpha$ and $\beta$:

$$
\begin{aligned}
& \alpha=\operatorname{arcctg} 8=\arcsin \frac{1}{\sqrt{65}}=\arccos \frac{8}{\sqrt{65}}=\frac{\pi}{2}-\arcsin \frac{8}{\sqrt{65}}=\ldots \\
& \beta=\operatorname{arcctg} \frac{7}{4}=\arcsin \frac{4}{\sqrt{65}}=\arccos \frac{7}{\sqrt{65}}=\frac{\pi}{2}-\arcsin \frac{7}{\sqrt{65}}=\ldots
\end{aligned}
$$

For example, the answer can be rewritten as

$$
-\frac{1}{9}\left(\pi+\arcsin \frac{8}{\sqrt{65}}+\arcsin \frac{7}{\sqrt{65}}\right)
$$

For convenience in checking, an approximate value of the expression is provided in the answer. This, of course, was not required of the participants.

Comment 2. The answer can be simplified to a single inverse trigonometric function. Indeed, if $\alpha=\operatorname{arctg} \frac{1}{8}, \beta=\operatorname{arctg} \frac{4}{7}$, then

$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{1 / 8+4 / 7}{1-1 / 8 \cdot 4 / 7}=\frac{3}{4},
$$

from which, since $\alpha+\beta \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$, we get that the answer is $\frac{1}{9}\left(\operatorname{arctg} \frac{3}{4}-2 \pi\right)$. Such simplification was not required of the participants.

Solution. Divide both sides of the equation by $\sqrt{65}=\sqrt{1^{2}+8^{2}}=\sqrt{4^{2}+7^{2}}$. Let $\alpha=\operatorname{arctg} \frac{1}{8}, \beta=\operatorname{arctg} \frac{4}{7}$; then $\sin \alpha=\frac{1}{\sqrt{65}}, \cos \alpha=\frac{8}{\sqrt{65}}, \sin \beta=\frac{4}{\sqrt{65}}, \cos \alpha=\frac{7}{\sqrt{65}}$, and the equation becomes

$$
\sin \alpha \sin x+\cos \alpha \cos x=\sin \beta \sin 8 x+\cos \beta \cos 8 x
$$

or $\cos (x-\alpha)=\cos (8 x-\beta)$. The solutions to this equation are the set

$$
\left[\begin{array}{l}
x-\alpha=8 x-\beta+2 \pi k, k \in \mathbb{Z} \\
x-\alpha=\beta-8 x+2 \pi n, n \in \mathbb{Z}
\end{array}\right.
$$

that is,

$$
\left[\begin{array}{l}
x=\frac{\beta-\alpha-2 \pi k}{7}, k \in \mathbb{Z} \\
x=\frac{\alpha+\beta+2 \pi n}{9}, n \in \mathbb{Z}
\end{array}\right.
$$

Since $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, the smallest negative solution is

$$
x_{1}=\frac{\beta-\alpha-2 \pi}{7}, \quad x_{2}=\frac{\alpha+\beta-2 \pi}{9}
$$

Since $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, we have

$$
(2 \pi-\alpha-\beta) \cdot \frac{1}{7} > (2 \pi-\alpha-\beta) \cdot \frac{1}{9} > (2 \pi-\alpha-\beta) \cdot \frac{1}{9} > (2 \pi-\alpha-\beta) \cdot \frac{1}{9}=-x_{2}
$$

(each of the factors on the left is greater than the corresponding factor on the right and all four are positive), from which $x_{1}<x_{2}<0$.