Task 6. Find the greatest negative root of the equation

$$
4 \sin (3 x)+13 \cos (3 x)=8 \sin (x)+11 \cos (x)
$$
Answer: $\frac{\alpha-\beta}{2} \approx-0.1651$, where $\alpha=\operatorname{arctg} \frac{4}{13}, \beta=\operatorname{arctg} \frac{8}{11}$.

Comment 1. Depending on the choice of the solution method and the inverse trigonometric function, the answer may have a different form. Let's provide several ways to describe $\alpha$ and $\beta$:

$$
\begin{aligned}
& \alpha=\operatorname{arcctg} \frac{13}{4}=\arcsin \frac{4}{\sqrt{185}}=\arccos \frac{13}{\sqrt{185}}=\frac{\pi}{2}-\arcsin \frac{13}{\sqrt{185}}=\ldots \\
& \beta=\operatorname{arcctg} \frac{11}{8}=\arcsin \frac{8}{\sqrt{185}}=\arccos \frac{11}{\sqrt{185}}=\frac{\pi}{2}-\arcsin \frac{11}{\sqrt{185}}=\ldots
\end{aligned}
$$

For example, the answer can be rewritten as

$$
\frac{1}{2}\left(\frac{\pi}{2}-\arcsin \frac{13}{\sqrt{185}}-\arcsin \frac{8}{\sqrt{185}}\right)
$$

For convenience of verification, 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 one inverse trigonometric function. Indeed, if $\alpha=\operatorname{arctg} \frac{4}{13}, \beta=\operatorname{arctg} \frac{8}{11}$, then

$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{4 / 13-8 / 11}{1+4 / 13 \cdot 8 / 11}=-\frac{12}{35}
$$

from which, since $\alpha-\beta \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$, we get that the answer is $-\frac{1}{2} \operatorname{arctg} \frac{12}{35}$. Such a simplification was not required of the participants.

Solution. Divide both sides of the equation by $\sqrt{185}=\sqrt{4^{2}+13^{2}}=\sqrt{8^{2}+11^{2}}$. Let $\alpha=\operatorname{arctg} \frac{4}{13}, \beta=\operatorname{arctg} \frac{8}{11}$; then $\sin \alpha=\frac{4}{\sqrt{185}}, \cos \alpha=\frac{13}{\sqrt{185}}, \sin \beta=\frac{8}{\sqrt{185}}, \cos \alpha=\frac{11}{\sqrt{185}}$, and the equation becomes

$$
\sin \alpha \sin 3 x+\cos \alpha \cos 3 x=\sin \beta \sin x+\cos \beta \cos x
$$

or $\cos (3 x-\alpha)=\cos (x-\beta)$. The solutions to this equation are the set

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

that is,

$$
\left[\begin{array}{l}
x=\frac{\alpha-\beta+2 \pi k}{2}, k \in \mathbb{Z} \\
x=\frac{\alpha+\beta^{2}+2 \pi n}{4}, n \in \mathbb{Z}
\end{array}\right.
$$

Since $0<\alpha<\beta<\frac{\pi}{2}$, the largest negative roots of the series in the set are the numbers $x_{1}=\frac{\alpha-\beta}{2}$ and $x_{2}=\frac{\alpha+\beta-2 \pi}{4}$. It remains to compare $x_{1}$ and $x_{2}$. For this, note that

$$
x_{2}=\frac{\alpha+\beta-2 \pi}{4}<\frac{\pi / 2+\pi / 2-2 \pi}{4}=-\frac{\pi}{4}=\frac{-\pi / 2}{2}<\frac{\alpha-\beta}{2}=x_{1} \text {. }
$$