Why does 0 = 1? Cos 0 = 1

It's equivalent to solving for $\displaystyle \prod_{r=0}^3 \left [ 1 - 4\sin(3 \cdot 5^r x) \sin(5^r x) \right ] = 1$ .

For the first term being multiplied, we apply the double angle formula and product to sum formula, and let $y=\cos x$ for simplicity sake.

$\begin{aligned} 1 - 4\sin(3x) \sin(x) &=& 1 + 2(-2\sin(3x) \sin(x)) \\ &=& 1 + 2(\cos(4x) - \cos(2x)) \\ &=& 1 - 2 \left [ 2(\cos^2(2x) - 1)^2 - 1 - \cos(2x) \right ] \\ &=& 1 + 2 \left [ 2((2y^2-1)^2 - 1)^2 - 1 - (2y^2-1) \right ] \\ &=& 16y^4 - 20y^2 + 5 \\ \end{aligned}$

Does the polynomial looks familiar? It's a slightly modified form of quintuple angle formula, $\cos(5A) = 16\cos^5 (A) - 20\cos^3(A) + 5\cos(A)$ , which can be easily derived from Chebyshev Polynomials.

Thus, $1 - 4\sin(3x) \sin(x) = \frac{ \cos(5x)}{\cos(x)}$ . The product in question can be stated as such:

$\begin{aligned} \frac{ \cos(5x)}{\cos(x)} \cdot \frac{ \cos(25x)}{\cos(5x)} \cdot \frac{ \cos(125x)}{\cos(25x)} \cdot \frac{ \cos(625x)}{\cos(125x)} &=& 1 \\ \Rightarrow \frac{ \cancel{\cos(5x)}}{\cos(x)} \cdot \frac{ \cancel{\cos(25x)}}{\cancel{\cos(5x)}} \cdot \frac{ \cancel{\cos(125x)}}{\cancel{\cos(25x)}} \cdot \frac{\cos(625x)}{\cancel{\cos(125x)}} &=& 1 \\ \end{aligned}$

Or $\cos(625x) = \cos(x) \Rightarrow \cos(625x) - \cos(x) = 0$

$\Rightarrow -2 \sin \left(\frac{625+1}2 x \right) \sin \left(\frac{625-1}2 x \right) = 0 \Rightarrow \sin(313x) = 0, \sin(312x) = 0$ .

Since $313$ is a prime number (which can be easily verified), all the values of $x$ that satisfy $\sin(313x) = 0$ falls in the given range are irreducible:

$\left \{ \frac{\pi}{313}, \frac{2\pi}{313}, \frac{3\pi}{313}, \ldots , \frac{312\pi}{313} \right \}$

Which gives a total of $312$ solutions, similarly, $\sin(312x) = 0$ gives a total of $311$ solutions. Hence totaling of $312+311 =\boxed{623}$ solutions.