We are given the equation $\cos(4x) = \sin(x)$ and we need to solve it in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We are also asked to justify some trigonometric identities and use them to solve a polynomial equation, and finally, deduce the value of $\sin(\frac{\pi}{10})$.

1. Problem Description

We are given the equation $\cos(4x) = \sin(x)$ and we need to solve it in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We are also asked to justify some trigonometric identities and use them to solve a polynomial equation, and finally, deduce the value of $\sin(\frac{\pi}{10})$.

2. Solution Steps

1. Solve $\cos(4x) = \sin(x)$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin(x) = \cos(\frac{\pi}{2} - x)$. So, we can rewrite the equation as:

$\cos(4x) = \cos(\frac{\pi}{2} - x)$.

This means that $4x = \frac{\pi}{2} - x + 2k\pi$ or $4x = -(\frac{\pi}{2} - x) + 2k\pi$, where $k$ is an integer.

Case 1: $4x = \frac{\pi}{2} - x + 2k\pi$

$5x = \frac{\pi}{2} + 2k\pi$

$x = \frac{\pi}{10} + \frac{2k\pi}{5}$

When $k = -1$, $x = \frac{\pi}{10} - \frac{2\pi}{5} = \frac{\pi}{10} - \frac{4\pi}{10} = -\frac{3\pi}{10}$.

When $k = 0$, $x = \frac{\pi}{10}$.

When $k = 1$, $x = \frac{\pi}{10} + \frac{2\pi}{5} = \frac{\pi}{10} + \frac{4\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2}$.

Case 2: $4x = -(\frac{\pi}{2} - x) + 2k\pi$

$4x = -\frac{\pi}{2} + x + 2k\pi$

$3x = -\frac{\pi}{2} + 2k\pi$

$x = -\frac{\pi}{6} + \frac{2k\pi}{3}$

When $k = 0$, $x = -\frac{\pi}{6}$.

When $k = 1$, $x = -\frac{\pi}{6} + \frac{2\pi}{3} = -\frac{\pi}{6} + \frac{4\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.

When $k = -1$, $x = -\frac{\pi}{6} - \frac{2\pi}{3} = -\frac{\pi}{6} - \frac{4\pi}{6} = -\frac{5\pi}{6}$ (not in the interval).

Therefore, the solutions are $x = -\frac{3\pi}{10}, -\frac{\pi}{6}, \frac{\pi}{10}, \frac{\pi}{2}$.

2. Represent the solutions on the trigonometric circle. (Skipped)

3. Justify the equalities:

a) $\cos(4x) = 1 - 2\sin^2(2x)$.

We know that $\cos(2\theta) = 1 - 2\sin^2(\theta)$. Let $\theta = 2x$, then $\cos(4x) = 1 - 2\sin^2(2x)$.

b) $\cos(4x) = 8\sin^4(x) - 8\sin^2(x) + 1$.

We know that $\cos(4x) = 1 - 2\sin^2(2x)$. We also know that $\sin(2x) = 2\sin(x)\cos(x)$.

So, $\cos(4x) = 1 - 2(2\sin(x)\cos(x))^2 = 1 - 2(4\sin^2(x)\cos^2(x)) = 1 - 8\sin^2(x)\cos^2(x)$.

Since $\cos^2(x) = 1 - \sin^2(x)$, we have

$\cos(4x) = 1 - 8\sin^2(x)(1 - \sin^2(x)) = 1 - 8\sin^2(x) + 8\sin^4(x)$.

Thus, $\cos(4x) = 8\sin^4(x) - 8\sin^2(x) + 1$.

4. Deduce that if $x = \sin(x)$, then $8X^4 - 8X^2 - X + 1 = 0$.

If $\cos(4x) = \sin(x)$ and $X = \sin(x)$, then $\cos(4x) = 8\sin^4(x) - 8\sin^2(x) + 1$ becomes $X = 8X^4 - 8X^2 + 1$. So, $8X^4 - 8X^2 - X + 1 = 0$.

5. Show that $1/2$ is a solution of $8X^4 - 8X^2 - X + 1 = 0$.

If $X=1/2$, $8(1/2)^4 - 8(1/2)^2 - (1/2) + 1 = 8(1/16) - 8(1/4) - 1/2 + 1 = 1/2 - 2 - 1/2 + 1 = -1 \ne 0$. So $1/2$ is NOT a solution.

Show that $1$ is a solution of $8X^4 - 8X^2 - X + 1 = 0$.

If $X=1$, $8(1)^4 - 8(1)^2 - (1) + 1 = 8 - 8 - 1 + 1 = 0$. So, $1$ is a solution.

Given $8X^4 - 8X^2 - X + 1 = (X-1)(2X+1)(aX^2 + bX + c)$. Expanding the right side gives:

$(X-1)(2X+1)(aX^2 + bX + c) = (2X^2 - X - 1)(aX^2 + bX + c) = 2aX^4 + (2b-a)X^3 + (2c-b-a)X^2 + (-c-b)X - c$.

Comparing coefficients with $8X^4 - 8X^2 - X + 1$, we get:

$2a = 8 \implies a = 4$

$2b - a = 0 \implies 2b = 4 \implies b = 2$

$2c - b - a = -8 \implies 2c - 2 - 4 = -8 \implies 2c = -2 \implies c = -1$

$-c - b = -1 \implies -(-1) - 2 = -1 \implies -1 = -1$

$-c = 1 \implies -(-1) = 1 \implies 1 = 1$

So, $8X^4 - 8X^2 - X + 1 = (X-1)(2X+1)(4X^2 + 2X - 1)$.

The roots are $X=1, X = -1/2$. Also $4X^2 + 2X - 1 = 0$.

$X = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$

Since we want $\sin(\frac{\pi}{10})$, it has to be positive. So, $X = \frac{-1 + \sqrt{5}}{4}$.

Then $\sin(\frac{\pi}{10}) = \frac{\sqrt{5}-1}{4}$.

3. Final Answer

The solutions to $\cos(4x) = \sin(x)$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are $x = -\frac{3\pi}{10}, -\frac{\pi}{6}, \frac{\pi}{10}, \frac{\pi}{2}$.

$\sin(\frac{\pi}{10}) = \frac{\sqrt{5}-1}{4}$