407. Solve the equation:

$$
\sqrt{1-x^{2}}=4 x^{3}-3 x
$$
$\triangle$ The domain of the equation is the interval $[-1 ; 1]$.

Let's introduce a trigonometric substitution $x=\cos \alpha$, where $\alpha \in[0 ; \pi]$. Then

$$
\sqrt{1-x^{2}}=\sqrt{1-\cos ^{2} \alpha}=\sin \alpha, \quad 4 x^{3}-3 x=4 \cos ^{3} \alpha-3 \cos \alpha=\cos 3 \alpha
$$

(here, in the first equality, we used the non-negativity of the sine on the interval $[0 ; \pi]$, and in the second - the formula for the cosine of a triple argument).

We obtain:

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

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

From this

$$
\alpha-\frac{\pi}{4}=\frac{\pi}{2}+\pi k, \quad \alpha=\frac{3 \pi}{4}+\pi k(k \in \mathrm{Z})
$$

or $2 \alpha-\frac{\pi}{4}=\pi n, \quad \alpha=\frac{\pi}{8}+\frac{\pi n}{2}(n \in \mathrm{Z})$.

Since $\alpha \in[0 ; \pi]$, $k$ and $n$ are not any integers. Only $k=0, n=0$ and $n=1$ are suitable. Therefore,

$$
\alpha_{1}=\frac{3 \pi}{4}, \quad \alpha_{2}=\frac{\pi}{8}, \quad \alpha_{3}=\frac{5 \pi}{8}
$$

134

Now we find the roots of the original equation:

$$
\begin{aligned}
x_{1}= & \cos \frac{3 \pi}{4}=-\frac{\sqrt{2}}{2}, \quad x_{2}=\cos \frac{\pi}{8}=\sqrt{\frac{1+\cos \frac{\pi}{4}}{2}}=\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}=\frac{1}{2} \sqrt{2+\sqrt{2}}, \\
x_{3}= & \cos \frac{5 \pi}{8}=-\frac{1}{2} \sqrt{2-\sqrt{2}} \\
& \text { Answer: }-\sqrt{2} / 2,(\sqrt{2+\sqrt{2}}) / 2,-(\sqrt{2-\sqrt{2}}) / 2 .
\end{aligned}
$$