The problem is divided into three parts: 1. Show that $cos(5x) = cos(x)(16cos^4(x) - 20cos^2(x) + 5)$ and $(1-cos(x))(4cos^2(x) + 2cos(x) - 1)^2 = 1 - cos(5x)$.

1. Problem Description

The problem is divided into three parts:

1. Show that $cos(5x) = cos(x)(16cos^4(x) - 20cos^2(x) + 5)$ and $(1-cos(x))(4cos^2(x) + 2cos(x) - 1)^2 = 1 - cos(5x)$.

2. Solve the equation $cos(5x) = 1$ for $x$ in $R$.

3. Deduce the values of $cos(\frac{2\pi}{5})$, $cos(\frac{4\pi}{5})$ and $cos(\frac{6\pi}{5})$ from the previous questions.

2. Solution Steps

Part 1:

We need to prove $cos(5x) = cos(x)(16cos^4(x) - 20cos^2(x) + 5)$.

Using the multiple angle formula, we have:

$cos(5x) = 16cos^5(x) - 20cos^3(x) + 5cos(x)$

$cos(5x) = cos(x)(16cos^4(x) - 20cos^2(x) + 5)$

Thus, the first statement is proven.

Next, we need to prove $(1-cos(x))(4cos^2(x) + 2cos(x) - 1)^2 = 1 - cos(5x)$.

Let $t = cos(x)$. We want to show that $(1-t)(4t^2+2t-1)^2 = 1 - (16t^5 - 20t^3 + 5t)$

$(1-t)(4t^2+2t-1)^2 = (1-t)(16t^4 + 16t^3 -4t^2+4t^2-4t+1) = (1-t)(16t^4 + 16t^3 - 8t^2-4t+1)$.

$(1-t)(16t^4 + 16t^3 - 8t^2-4t+1) = 16t^4+8t^3-4t^2-4t+1 + 32t^3= 16t^4+16t^3-16^3t^2 -4t+1 -16t^5 - 16t^4+8t^3 -t =-16t^5-16t^5 - 4t +1 + -16t^4+8t^3 -t =1$

$(1-t)(16t^4+16t^3-8t^2-4t+1) = (1-t)(16t^4+16t^3-4t^2-1)+1$

$16t^4+16t^3-8t^2-4t+1 - (16t^5+16t^4 -8t^3 -4t^2+t)= 16t^4-1$

$1 -1t6t^5+(16-16)t^4 (16+t)-{t} (-8+8)t + (-4+t) t+1 =1-16t^5t-2(1))=-5t$.

Part 2:

We want to solve $cos(5x) = 1$ for $x$ in $R$.

$cos(5x) = 1$ implies $5x = 2k\pi$, where $k$ is an integer.

So, $x = \frac{2k\pi}{5}$, $k \in Z$.

Part 3:

We need to deduce the values of $cos(\frac{2\pi}{5})$, $cos(\frac{4\pi}{5})$ and $cos(\frac{6\pi}{5})$ from the previous questions.

From part 1, we have $cos(5x) = cos(x)(16cos^4(x) - 20cos^2(x) + 5)$. If $cos(5x) = 1$, then $x = \frac{2k\pi}{5}$, so

$cos(\frac{10k\pi}{5}) = 1$, then $k = 0, 1, 2, 3, 4$, $x = 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}.$

Then we have $(1-cos(x))(4cos^2(x) + 2cos(x) - 1)^2 = 1-cos(5x) = 0$.

So $1 - cos(x) = 0$ or $(4cos^2(x) + 2cos(x) - 1)^2 = 0$

$cos(x) = 1$ or $4cos^2(x) + 2cos(x) - 1 = 0$.

If $cos(x) = 1$, then $x = 0$.

Consider the equation $4cos^2(x) + 2cos(x) - 1 = 0$.

Let $y = cos(x)$. Then $4y^2 + 2y - 1 = 0$.

Using the quadratic formula, we get $y = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.

So $cos(x) = \frac{-1+\sqrt{5}}{4}$ or $cos(x) = \frac{-1-\sqrt{5}}{4}$.

$cos(\frac{2\pi}{5}), cos(\frac{4\pi}{5}), cos(\frac{6\pi}{5})$ are solutions of $cos(x) = \frac{-1+\sqrt{5}}{4}$ or $cos(x) = \frac{-1-\sqrt{5}}{4}$ and none of them are equal to $cos(0)=1$.

Note that $cos(\frac{2\pi}{5})>0$ since $\frac{2\pi}{5}$ is in the first quadrant.

$cos(\frac{4\pi}{5})<0$ and $cos(\frac{6\pi}{5})<0$ since they are in the second and third quadrants respectively.

$cos(\frac{2\pi}{5}) = \frac{\sqrt{5}-1}{4}$.

$cos(\frac{4\pi}{5}) = \frac{-\sqrt{5}-1}{4}$.

$cos(\frac{6\pi}{5}) = cos(2\pi-\frac{4\pi}{5}) = cos(\frac{4\pi}{5}) = \frac{-\sqrt{5}-1}{4}$.

3. Final Answer

$cos(\frac{2\pi}{5}) = \frac{\sqrt{5}-1}{4}$

$cos(\frac{4\pi}{5}) = \frac{-\sqrt{5}-1}{4}$

$cos(\frac{6\pi}{5}) = \frac{-\sqrt{5}-1}{4}$