We are tasked with using Newton's method to approximate the root of the equation:

$5 \sin(x) = x$

Newton's method uses the iterative formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

where $f(x) = 5 \sin(x) - x$, and $f'(x)$ is its derivative. Let's find the derivative of $f(x)$:

$f'(x) = 5 \cos(x) - 1$

Given that the initial approximation is $x_1 = 1$, we will now calculate the second approximation $x_2$ and the third approximation $x_3$.

1. Calculate $f(x_1)$ and $f'(x_1)$:

   • $f(x_1) = 5 \sin(1) - 1$
   • $f'(x_1) = 5 \cos(1) - 1$

2. Apply Newton's method to get $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}$

3. Apply the same process to calculate $x_3$: $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}$

Let me compute these approximations for you.Using Newton's method, the second and third approximations for the root of the equation $5 \sin(x) = x$ are:

• The second approximation, $x_2 \approx -0.885$
• The third approximation, $x_3 \approx 0.493$

Would you like more details or have any questions on the process?

Here are some related questions to explore further:

1. How does Newton's method converge to a root?
2. What are the advantages of Newton's method compared to other root-finding methods?
3. Can Newton's method fail to converge? Under what conditions?
4. How do we interpret the physical meaning of the equation $5 \sin(x) = x$?
5. How sensitive is Newton's method to the choice of the initial guess?

Tip: Always check the derivative when using Newton's method—if it's near zero, the method may fail due to division by very small numbers.