using newtons method, find the second and third approximation of 3sinx = 3 where x1 = 1.

We are tasked with applying **Newton's method** to find the second and third approximations for the equation \( 3\sin(x) = 3 \), starting from the initial approximation \( x_1 = 1 \).

### Step 1: Rearrange the equation
First, rewrite the given equation in a standard form suitable for Newton's method, where we need the equation in the form \( f(x) = 0 \). 

\[
3\sin(x) = 3 \implies \sin(x) = 1
\]

Now define the function \( f(x) \) as:

\[
f(x) = \sin(x) - 1
\]

### Step 2: Newton's Method Formula
Newton's method iteratively improves the approximation to the root by using the formula:

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

Where:
- \( f(x) = \sin(x) - 1 \)
- \( f'(x) = \cos(x) \) (the derivative of \( f(x) \))

### Step 3: Calculate the second approximation \( x_2 \)
We are given that \( x_1 = 1 \).

1. Compute \( f(x_1) \):

\[
f(x_1) = \sin(1) - 1 \approx 0.84147 - 1 = -0.15853
\]

2. Compute \( f'(x_1) \):

\[
f'(x_1) = \cos(1) \approx 0.5403
\]

3. Use Newton's formula to find \( x_2 \):

\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1 - \frac{-0.15853}{0.5403} \approx 1 + 0.29346 \approx 1.2935
\]

### Step 4: Calculate the third approximation \( x_3 \)
Now, repeat the process using \( x_2 = 1.2935 \).

1. Compute \( f(x_2) \):

\[
f(x_2) = \sin(1.2935) - 1 \approx 0.96136 - 1 = -0.03864
\]

2. Compute \( f'(x_2) \):

\[
f'(x_2) = \cos(1.2935) \approx 0.2756
\]

3. Use Newton's formula to find \( x_3 \):

\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.2935 - \frac{-0.03864}{0.2756} \approx 1.2935 + 0.1402 \approx 1.4337
\]

### Final Result:
- Second approximation \( x_2 \approx 1.2935 \)
- Third approximation \( x_3 \approx 1.4337 \)

Would you like more details or have any questions?

### 5 Related Questions:
1. How does Newton's method converge to the root of a function?
2. What happens if the derivative \( f'(x) \) is close to zero in Newton's method?
3. Can Newton's method fail? If so, under what conditions?
4. How does Newton's method compare to other root-finding algorithms like bisection or secant method?
5. What are the limitations of using Newton's method for non-linear equations?

### Tip:
Newton's method is very sensitive to the initial guess. A poor choice of starting point may result in divergence or slow convergence.

Using Newton's method, find the second and third approximation of 3sinx = 3 where x1 = 1.

Learn how to apply Newton's method to find the second and third approximations for the equation 3sin(x) = 3, starting with an initial guess of x1 = 1. This solution walks through the steps of calculating the next approximations using derivatives and trigonometric identities. Suitable for advanced high school or college-level calculus students.

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