[ Trigonometric substitutions $]$

How many roots does the equation $8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ have on the interval $[0,1]$?
Note that $8 x^{4}-8 x^{2}+1=2\left(2 x^{2}-1\right)^{2}-1$. By making the substitution $x=\cos \varphi$, we get $8 \cos \varphi \cos 2 \varphi \cos 4 \varphi=-1$. Multiplying by $\sin \varphi$, we obtain $\sin 8 \varphi=-\sin \varphi$, from which $8 \varphi=-\varphi+2 k \pi$ or $8 \varphi=\pi+\varphi+2 k \pi$, that is, $x=\cos 2 k \pi / 9$ or $x=\cos (\pi / 7+2 k \pi / 7)$. On the interval $[0,1]$, there are four roots of the equation: $\cos 2 \pi / 9, \cos 4 \pi / 9, \cos \pi / 7$ and $\cos 3 \pi / 7$ (the root $x=1-$ is extraneous, it arose from the multiplication by $\sin \varphi)$.

## Answer

Four roots.