Cosine equation

Note that $8x(2x^2 - 1)(8x^4 - 8x^2 + 1) - 1 \; = \; (2x-1)(8x^3 - 6x - 1)(8x^3 + 4x^2 - 4x - 1) \; = \; (2x-1)f(2x)g(2x)$ where $f(x) \; = \; x^3 - 3x - 1 \hspace{2cm} g(x) \; = \; x^3 + x^2 - 2x - 1$ are integer polynomials. Since $f(x+1) \; = \; x^3 + 3x^2 - 3 \hspace{2cm} g(x+2) \; = \; x^3 + 7x^2 + 14x + 7$ we deduce from Eisenstein's Irreducibility Criterion that $f(x)$ ( $p=3$ ) and $g(x)$ ( $p=7$ ) are irreducible over the rationals, and hence certainly have no rational zeros. Thus the only rational solution to the initial equation is $x=\boxed{\tfrac12}$ .

By rational root theorem , all the potential rational roots in the interval $[0,1]$ are $1, 1/2, 1/4, 1/8$ . Trial and error shows that $1/2 = \boxed{0.5}$ is the only such rational root.