Solve the following equation on the set of integers:

$$
\sin \left(\frac{\pi}{3}\left(x-\sqrt{x^{2}-3 x-12}\right)\right)=0
$$
Solution. $\sin \alpha$ is exactly 0 if $\alpha$ is an integer multiple of $\pi$, so $x$ will be a solution to the equation if $x-\sqrt{x^{2}-3 x-12}$ is an integer divisible by 3. Due to divisibility, $\sqrt{x^{2}-3 x-12}$ must also be an integer, let's denote this by $a$ $(a \geq 0)$.

Then

$$
x^{2}-3 x-12=a^{2}
$$

Multiplying by 4 and completing the square:

$$
4 x^{2}-12 x-48=4 a^{2}, \quad(2 x-3)^{2}-4 a^{2}=57
$$

Factoring:

$$
(2 x-3+2 a)(2 x-3-2 a)=57
$$

Since $a \geq 0,2 x-3+2 a \geq 2 x-3-2 a$. The prime factorization of 57 is: $57=3 \cdot 19$, so the following cases are possible:

| $2 x-3+2 a$ | $2 x-3-2 a$ | $4 a$ | $a$ | $x$ | $x-a$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 57 | 1 | 56 | 14 | 16 | 2 |
| 19 | 3 | 16 | 4 | 7 | 3 |
| -1 | -57 | 56 | 14 | -13 | -27 |
| -3 | -19 | 16 | 4 | -4 | 8 |

From the table, we can see that $x-a$ will be divisible by 3 when $x=7$, or $x=-13$.

By substitution, we can verify that these are indeed solutions to the equation.