2. Using a suitable substitution, determine the number of roots of the equation

$$
8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1
$$

that lie within the interval $[0,1]$.
2. We immediately see that $x=0$ and $x=1$ are not solutions, so we introduce the substitution $x=\cos \varphi$, with $\varphi \in\left(0, \frac{\pi}{2}\right)$. The equation becomes

$$
8 \cos \varphi\left(1-2 \cos ^{2} \varphi\right)\left(8 \cos ^{4} \varphi-8 \cos ^{2} \varphi+1\right)=1
$$

5 points

or, because $\cos 2 \varphi=2 \cos ^{2} \varphi-1$ and $\cos 4 \varphi=8 \cos ^{4} \varphi-8 \cos ^{2} \varphi+1$,

$$
-8 \cos \varphi \cos 2 \varphi \cos 4 \varphi=1
$$

Multiply both sides of the equation by $\sin \varphi$. (The resulting equation is equivalent to the previous one since $\sin \varphi \neq 0$ for $\varphi \in\left(0, \frac{\pi}{2}\right)$.)

$$
\begin{gathered}
-4 \sin 2 \varphi \cos 2 \varphi \cos 4 \varphi=\sin \varphi \\
-2 \sin 4 \varphi \cos 4 \varphi=\sin \varphi \\
\sin 8 \varphi+\sin \varphi=0 \\
2 \sin \frac{9 \varphi}{2} \cos \frac{7 \varphi}{2}=0
\end{gathered}
$$

so the solutions are $\varphi=\frac{2}{9} k \pi$ and $\varphi=\frac{\pi}{7}+\frac{2}{7} k \pi$.

5 points

In the interval $\left(0, \frac{\pi}{2}\right)$, the solutions are $\varphi=\frac{2 \pi}{9}, \varphi=\frac{4 \pi}{9}, \varphi=\frac{\pi}{7}$, and

$\varphi=\frac{3 \pi}{7}$, so the original equation has 4 solutions in the interval $[0,1]$.

5 points