To solve this problem, we will use the Taylor Remainder Theorem.

Step 1: Taylor Expansion for $\cos(x)$

The Taylor series expansion of $\cos(x)$ at $a = 0$ is:

$f(x) = \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$

The degree-5 Taylor polynomial is:

$T_5(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24}.$

The remainder $R_5(x)$ is given by the next term in the series:

$R_5(x) = \frac{f^{(6)}(c)}{6!} x^6,$

where $c$ is some value between $0$ and $x$, and $f^{(6)}(x) = -\cos(x)$.

Step 2: Bound on the Remainder

For $|x| \leq 1$, the maximum value of $|-\cos(c)|$ is $1$. Thus, the magnitude of the remainder is:

$|R_5(x)| \leq \frac{1}{6!} |x^6| = \frac{1}{720} |x^6|.$

We want $|R_5(x)| \leq 0.002044$. This gives:

$\frac{1}{720} |x^6| \leq 0.002044.$

Step 3: Solve for $|x|$

Multiply through by $720$:

$|x^6| \leq 0.002044 \times 720 = 1.472.$

Take the 6th root of both sides:

$|x| \leq \sqrt[6]{1.472}.$

Step 4: Approximation of $\sqrt[6]{1.472}$

Using numerical approximation:

$\sqrt[6]{1.472} \approx 0.874.$

Thus, the values of $|x|$ for which the approximation is within $0.002044$ of the true value are:

$|x| \leq 0.874.$

Final Answer:

$|x| \leq 0.874.$

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Would you like further clarification or a detailed breakdown of the calculations?
Here are 5 related questions to expand your understanding:

1. How does the Taylor Remainder Theorem estimate the error in approximations?
2. Why is $f^{(6)}(c) = -\cos(c)$ relevant in this case?
3. What are the implications of using a higher-degree Taylor polynomial for $\cos(x)$?
4. How does the interval $|x| \leq 1$ simplify the remainder calculation?
5. What are the applications of Taylor series in physics or engineering?

Tip: Practice estimating the remainder for different degrees to better understand convergence!