5. Let's make the substitution of variables:

$$
3^{2 x^{2}-3 x+1}=u, \alpha=\pi x+\frac{7 \pi n}{6}
$$

The equation can be transformed into:

$$
\log _{2}\left(u^{2}-6 u+17\right)=\frac{6\left(2 \log _{2} \sin \alpha+1\right)}{\left(2 \log _{2} \sin \alpha+1\right)^{2}+1}
$$

Now let's introduce the variable $t: t=2 \log _{2} \sin \alpha+1$. Then the right-hand side of the equation can be transformed into:

$$
g(t)=\frac{6 t}{t^{2}+1}
$$

The function $g(t)$ is negative for negative values of the argument, and for positive values, it can be represented as:

$$
g(t)=\frac{6}{t+\frac{1}{t}}
$$

from which it is clear that the function takes its maximum value when the denominator is positive and minimal. This occurs at $t=1$, i.e., when $\alpha=\frac{\pi}{2}+2 \pi k$. At this point, the maximum value of the right-hand side of the equation is 3. The left-hand side of the equation

$$
f(u)=\log _{2}\left(u^{2}-6 u+17\right)=\log _{2}\left((u-3)^{2}+8\right) \geqslant 3
$$

is always greater than or equal to 3 and achieves its minimum value at $u=3$. From this, we can find the values of the variable $x$:

$$
3^{2 x^{2}-3 x+1}=3 \Rightarrow 2 x^{2}-3 x+1=1 \Rightarrow x=0 \cup x=\frac{3}{2}
$$

which are candidates for being the roots of the original equation. The values of the variable $x$ on the left and right sides must coincide, so the solutions will be for such values of $n$ that at least one of the conditions is satisfied:

$$
\pi \cdot 0+\frac{7 \pi n}{6}=\frac{\pi}{2}+2 \pi k \text { or } \pi \cdot \frac{3}{2}+\frac{7 \pi n}{6}=\frac{\pi}{2}+2 \pi k
$$

In both cases, we get linear Diophantine equations, which are solved by representing $k$ through classes of divisibility by 7 with a remainder $k=7 l+r,(r=0, \ldots, 6)$. The first of these equations in terms of the variable $n$ reduces to the equation $n=\frac{3+12 k}{7}$, which has a unique solution $n=9$ on the given interval of natural numbers. The second equation reduces to the equation: $n=\frac{12 k-6}{7}$, which has a unique solution $n=6$.
Answer: $\{6,9\}$.

## Answers to option $6-2$