# Solution.

Let's rewrite the equation as

$$
\cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)+\sin \left(2 \pi-\frac{\pi x^{2}}{6}\right) \sin \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)=\frac{p+3}{2} \text {, or }
$$

$\cos \left(2 \pi-\frac{\pi x^{2}}{6}-\frac{\pi}{3} \sqrt{9-x^{2}}\right)=\frac{p+3}{2}$.

Consider the expression $t=2 \pi-\frac{\pi x^{2}}{6}-\frac{\pi}{3} \sqrt{9-x^{2}}$ separately, which can be rewritten as:

$$
\begin{aligned}
2 \pi-\frac{\pi x^{2}}{6}- & \frac{\pi}{3} \sqrt{9-x^{2}}=\frac{\pi}{6}\left(12-x^{2}-2 \sqrt{9-x^{2}}\right)=\frac{\pi}{6}\left(9-x^{2}-2 \sqrt{9-x^{2}}+3\right) \\
= & \frac{\pi}{6}\left(\left(\sqrt{9-x^{2}}-1\right)^{2}+2\right)
\end{aligned}
$$

Considering that the root takes values from 0 to 3, then $t \in\left[\frac{\pi}{3} ; \pi\right]$, so $\cos t \in$ $\left[-1 ; \frac{1}{2}\right]$. Let's find the range in which the parameter varies: $-1 \leq \frac{p+3}{2} \leq \frac{1}{2} \Rightarrow-5 \leq p \leq-2$

## Answer: -2
8. Specify the greatest value of the parameter $p$ for which the equation has at least one solution

$2 \cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)-3=p-2 \sin \left(-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)$. (16 points)