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For $x \ne k\pi$ for any integer $k$ we can multiply both sides by $128\sin(x).$ Then since $2\sin(ax)\cos(ax) = \sin(2ax)$ the product on the LHS of the equation "collapses" to $\sin(128x),$ leaving us with the equation $\sin(128x) = \sin(x)$ such that $\sin(x) \ne 0.$ Then the solutions will be either of the form

• (i) $128x = x + 2k\pi \Longrightarrow x = \dfrac{2k\pi}{127}$ or

• (ii) $128x = \pi - x + 2k\pi \Longrightarrow x = \dfrac{(2k + 1)\pi}{129}.$

For (i), on the given interval $[0,2\pi]$ we can have $k$ taking on any integer value from $1$ to $126,$ resulting in $126$ solutions.

For (ii), we can have $k$ taking on any integer value from $0$ to $128$ except for $k = 64,$ as this value would make $\sin(x) = 0.$ This results in $128$ more solutions.

Thus the desired number of solutions is $126 + 128 = \boxed{254}.$

L.H.S= $\cos{(2^0x)} \cos{(2^1x)} \cos{(2^2x)} \cdots \cos{(2^6x)}$

R.HS= ${(\frac{1}{2})}^7$

We know that $0\leq x \leq{2\pi}\implies -1\leq {\cos{ax}} \leq 1$ for any integer $a.$

With alitel observation, we can say L.H.S=R.H.S iff $\cos{(2^0x)} =\cos{(2^1x)} =\cos{(2^2x)}= \cdots =\cos{(2^6x)}=\frac{1}{2}$

For $\cos{(2^0x)}$ have periode $2\pi\implies 2=2^1$ solutions,

For $\cos{(2^1x)}$ have periode $4\pi\implies 4=2^2$ solutions,

For $\cos{(2^2x)}$ have periode $8\pi\implies 8=2^3$ solution,

For $\cos{(2^3x)}$ have periode $16\pi\implies 16=2^4$ solutions,

$\vdots$

For $\cos{(2^6x)}$ have periode $128\pi\implies 128=2^7$ solutions.

Number of solutions $=2^1+2^2+2^3+\cdots+2^7$ $=\frac{2(2^7-1)}{2-1}=\boxed{254}$