Solve the following equation:

$$
4 \mathrm{x}^{3}-3 x+\left(1-4 x^{2}\right) \sqrt{1-x^{2}}=0
$$
I. solution. Subtract the third term from both sides of the equation, and square it. This way, for $z=x^{2}$, we get a cubic equation, which, when rearranged, can be written as

$$
32 z^{3}-48 z^{2}+18 z-1=0
$$

Let us introduce the unknown $u=2 z$, then we get the equation

$$
4 u^{3}-12 u^{2}+9 u-1=0
$$

Here, the sum of the coefficients is zero, so $u=1$ is a root of the equation. Factoring out the $(u-1)$ root factor, we get

$$
(u-1)\left(4 u^{2}-8 u-1\right)=0
$$

The roots of the equation are

$$
u_{1}=1, \quad u_{2}=\frac{4+\sqrt{12}}{4} \text { and } u_{3}=\frac{4-\sqrt{12}}{4}
$$

thus

$$
z_{1}=\frac{1}{2}, \quad z_{2}=\frac{2+\sqrt{3}}{4}, \quad z_{3}=\frac{2-\sqrt{3}}{4}
$$

The possible solutions of the original equation are therefore:

$$
x_{1,2}= \pm \frac{1}{\sqrt{2}}, \quad x_{3,4} \frac{ \pm \sqrt{2+\sqrt{3}}}{2}, \quad x_{5,6}=\frac{ \pm \sqrt{2-\sqrt{3}}}{2}
$$

Since we also squared, the correctness of the obtained solutions must be verified. Performing this (somewhat tedious work) results in the solutions:

$$
x_{2}=-\frac{1}{\sqrt{2}}, \quad x_{3}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } x_{5}=\frac{\sqrt{2-\sqrt{3}}}{2}
$$

II. solution. The expression $\sqrt{1-x^{2}}$ only makes sense if $-1 \leqq x \leqq 1$; for every such $x$, there is an angle $\alpha$ between $-90^{\circ}$ and $90^{\circ}$ such that $x=\sin \alpha$. With this substitution, the equation becomes:

$$
4 \sin ^{3} \alpha-3 \sin \alpha+\left(1-4 \sin ^{2} \alpha\right) \cos \alpha=0
$$

(We used the fact that $\sqrt{1-\sin ^{2} \alpha}=\cos \alpha$, if $-90^{\circ} \leqq \alpha \leqq 90^{\circ}$.) It is known that $4 \sin ^{3} \alpha-3 \sin \alpha=-\sin 3 \alpha$ and $\left(1-4 \sin ^{2} \alpha\right) \cos \alpha=\left(4 \cos ^{2} \alpha-3\right) \cos \alpha=\cos 3 \alpha$. From this, it follows that $\sin 3 \alpha=\cos 3 \alpha$.

The solution to this is:

$$
3 \alpha=45^{\circ}+k \cdot 360^{\circ} \text { and } 3 \alpha=-135^{\circ}+k \cdot 360^{\circ} \text {. }
$$

The $\alpha$ values between $-90^{\circ}$ and $90^{\circ}$ are therefore:

$$
\alpha_{1}=15^{\circ}, \quad \alpha_{2}=75^{\circ}, \quad \alpha_{3}=45^{\circ}
$$

thus the solution to the original equation is:

$$
x_{1}=\sin 15^{\circ}=\frac{\sqrt{2-\sqrt{3}}}{2}, x_{2}=\sin 75^{\circ}=\frac{\sqrt{2+\sqrt{3}}}{2}, x_{3}=\sin \left(-45^{\circ}\right)=-\frac{1}{\sqrt{2}}
$$

Since we arrived at the solution using equivalent transformations, there is no need for substitution.