37. Find the smallest $a \in \mathbf{N}^{*}$, such that the following equation has real roots:
$$
\cos ^{2} \pi(a-x)-2 \cos \pi(a-x)+\cos \frac{3 \pi x}{2 a} \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+2=0 .
$$
$$
\begin{array}{l}
\text { 37. Original equation } \Leftrightarrow[\cos \pi(a-x)-1]^{2}+\left[\cos \frac{3 \pi}{2 a} x \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+1\right]=0 \\
\Leftrightarrow[\cos \pi(a-x)-1]^{2}+\frac{1}{2}\left[\cos \left(\frac{2 \pi}{a} x+\frac{\pi}{3}\right)+\cos \left(\frac{\pi x}{a}-\frac{\pi}{3}\right)+2\right]=0 \\
\Leftrightarrow\left\{\begin{array} { l } 
{ \cos \pi ( a - x ) = 1 } \\
{ \cos ( \frac { 2 \pi } { a } x + \frac { \pi } { 3 } ) = - 1 } \\
{ \cos ( \frac { \pi x } { a } - \frac { \pi } { 3 } ) = - 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a-x=2 k_{1} \cdots \cdots \text { (1) } \\
\frac{2 x}{a}+\frac{1}{3}=2 k_{2}-1, k_{i} \in \mathbf{Z} \cdots \text { (2) } \\
\frac{x}{a}-\frac{1}{3}=2 k_{3}-1 \cdots \cdots \text { (3) }
\end{array}\right.\right. \\
\Leftrightarrow\left\{\begin{array}{l}
a-x=2 k_{1} \cdots \cdots \text { (1) } \\
\left(3 k_{2}-2\right) \cdot a=3 x \cdots \cdots \text { (2) } \\
3 x-a=3 a\left(2 k_{3}-1\right) \cdots \cdots \text { (3) }
\end{array}\right.
\end{array}
$$

From (1), we know: since $a \in \mathbf{N}^{*}$, so $x \in \mathbf{Z}$ and $x, a$ have the same parity.
From (2), we know that $a$ is a multiple of 3.
If $x, a$ are both odd, substituting into (3) leads to a contradiction, so $x, a$ must both be even.
Therefore, $a$ is a multiple of 6, and since $a \in \mathbf{N}^{*}$, we have $a \geqslant 6$.
When $a=6$, $x=8$ is a real solution. Therefore, $a_{\text {min }}=6$.
Note 1: If the equation is changed to:
$$
\cos ^{2} \pi(a-x)+2 \cos \pi(a-x)+\cos \frac{3 \pi x}{2 a} \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+2=0
$$

Using the same method, we get $a_{\text {min }}=3$, at which point $x=4$ is a real solution.
Note 2: When simplifying $\cos \frac{3 \pi}{2 a} x \cdot \cos \left(\frac{\pi x}{2 a}+\frac{\pi}{3}\right)+1=0$, we can let $\theta=\frac{\pi x}{2 a}+\frac{\pi}{3}$, then $3 \theta=\frac{3 \pi x}{2 a}+\pi$.
Then the equation $\Leftrightarrow \cos (3 \theta-\pi) \cdot \cos \theta+1=0 \Leftrightarrow \cos 3 \theta \cos \theta-1=0 \Leftrightarrow\left(4 \cos ^{3} \theta-3 \cos \theta\right) \cos \theta-1=0 \Leftrightarrow\left(\cos ^{2} \theta-\right.$ 1) $\left(4 \cos ^{2} \theta+1\right)=0 \Leftrightarrow \cos ^{2} \theta=1$.
$$