Task B-3.1. (20 points) Find all real solutions of the equation

$$
4 x^{3}-\sqrt{1-x^{2}}-3 x=0
$$
First solution. Let's write the equation in the form $\sqrt{1-x^{2}}=4 x^{3}-3 x$.

For the equation to be defined, it must be $1-x^{2} \geq 0$ i.e., $x \in[-1,1]$. Therefore, we can set $x=\sin \alpha$, where $\alpha \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Now the equation becomes

$$
\sqrt{1-\sin ^{2} \alpha}=4 \sin ^{3} \alpha-3 \sin \alpha \quad \text { i.e., } \quad|\cos \alpha|=-\sin 3 \alpha
$$

For $\alpha \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\cos \alpha \geq 0$, so the equation takes the form

$$
\cos \alpha=-\sin 3 \alpha \quad \text { i.e., } \quad \sin 3 \alpha+\sin \left(\frac{\pi}{2}-\alpha\right)=0
$$

The left side can be written in the form of a product:

$$
\begin{gathered}
2 \sin \frac{3 \alpha+\frac{\pi}{2}-\alpha}{2} \cdot \cos \frac{3 \alpha-\frac{\pi}{2}+\alpha}{2}=0 \quad \text{i.e.,} \\
\sin \left(\alpha+\frac{\pi}{4}\right) \cdot \cos \left(2 \alpha-\frac{\pi}{4}\right)=0
\end{gathered}
$$

We have two possibilities:

$1^{\circ} \sin \left(\alpha+\frac{\pi}{4}\right)=0$

The general solution to this equation is $\alpha=-\frac{\pi}{4}+k \pi, k \in \mathbb{Z}$;

$2^{\circ} \cos \left(2 \alpha-\frac{\pi}{4}\right)=0$

whose general solution is $\alpha=\frac{3 \pi}{8}+\frac{k \pi}{2}, k \in \mathbb{Z}$.

The equation has three solutions in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]:-\frac{\pi}{4},-\frac{\pi}{8}, \frac{3 \pi}{8}$.

The solutions to the original equation are:

$x_{1}=\sin \left(-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}$,
$x_{2}=\sin \left(-\frac{\pi}{8}\right)=-\sqrt{\frac{1-\cos \frac{\pi}{4}}{2}}=-\frac{1}{2} \sqrt{2-\sqrt{2}}$

$x_{3}=\sin \left(\frac{3 \pi}{8}\right)=\sqrt{\frac{1-\cos \frac{3 \pi}{4}}{2}}=\frac{1}{2} \sqrt{2+\sqrt{2}}$.