8. Indicate the smallest value of the parameter $a$ for which the equation has at least one solution

$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)+1=a+2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\frac{\pi x^{2}}{12}\right)$.
Solution.

Rewrite the equation as

$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)-2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\pi-\frac{\pi x^{2}}{12}\right)=a-1$, or

$\sin \left(\pi-\frac{\pi x^{2}}{12}-\frac{\pi}{6} \sqrt{9-x^{2}}\right)=\frac{a-1}{2}$.

Consider the expression $t=\pi-\frac{\pi x^{2}}{12}-\frac{\pi}{6} \sqrt{9-x^{2}}$ separately, which can be rewritten as:

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

Considering that the root takes values from 0 to 3, then $t \in\left[\frac{\pi}{6} ; \frac{\pi}{2}\right]$, so $\sin t \in$ $\left[\frac{1}{2} ; 1\right]$. Let's find the range in which the parameter varies: $\frac{1}{2} \leq \frac{a-1}{2} \leq 1 \Rightarrow 2 \leq a \leq 3$.

Answer: 2