11.3. Solve in the set of real numbers the equation

$$
\cos \frac{\pi}{x}=-x^{6}-4 x^{5}+2 x^{4}+12 x^{3}-9 x^{2}-1
$$
Solution.

We have

$$
\begin{gathered}
1 \geq -\cos \frac{\pi}{x} = x^{6} + 4 x^{5} - 2 x^{4} - 12 x^{3} + 9 x^{2} + 1 = x^{2}\left(x^{4} + 4 x^{3} - 2 x^{2} - 12 x + 9\right) + 1 = \\
= x^{2}\left[\left(x^{2} + 2 x\right)^{2} - 6 x^{2} - 12 x + 9\right] + 1 = x^{2}\left[\left(x^{2} + 2 x\right)^{2} - 6\left(x^{2} + 2 x\right) + 9\right] + 1 = \\
= \left[x\left(x^{2} + 2 x - 3\right)\right]^{2} + 1 = [x(x-1)(x+3)]^{2} + 1 \geq 1
\end{gathered}
$$

which implies $[x(x-1)(x+3)]^{2} + 1 = 1$, i.e., $x=0, x=1$ or $x=-3$.

The value $x=0$ is not a solution because it is not in the domain of the function $f(x) = \cos \frac{\pi}{x}$.

For $x=1$, we have $-\cos \frac{\pi}{x} = -\cos \pi = 1$, so $x=1$ is a solution.

For $x=-3$, we have $-\cos \frac{\pi}{x} = -\cos \left(-\frac{\pi}{3}\right) = -\cos \frac{\pi}{3} = -\frac{1}{2} \neq 1$, so $x=-3$ is not a solution.

In conclusion, the only real solution to the equation in the statement is $x=1$.