Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{13 x^{3}}{100} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{13 x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 5}{- \frac{39 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{13 (3.0000000000)^{3}}{100} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{39 (3.0000000000)^{2}}{100} + \cos{\left((3.0000000000) \right)}} = 3.3624717173$ $LaTeX: x_{2} = (3.3624717173) - \frac{- \frac{13 (3.3624717173)^{3}}{100} + \sin{\left((3.3624717173) \right)} + 5}{- \frac{39 (3.3624717173)^{2}}{100} + \cos{\left((3.3624717173) \right)}} = 3.3325234059$ $LaTeX: x_{3} = (3.3325234059) - \frac{- \frac{13 (3.3325234059)^{3}}{100} + \sin{\left((3.3325234059) \right)} + 5}{- \frac{39 (3.3325234059)^{2}}{100} + \cos{\left((3.3325234059) \right)}} = 3.3323203593$ $LaTeX: x_{4} = (3.3323203593) - \frac{- \frac{13 (3.3323203593)^{3}}{100} + \sin{\left((3.3323203593) \right)} + 5}{- \frac{39 (3.3323203593)^{2}}{100} + \cos{\left((3.3323203593) \right)}} = 3.3323203500$ $LaTeX: x_{5} = (3.3323203500) - \frac{- \frac{13 (3.3323203500)^{3}}{100} + \sin{\left((3.3323203500) \right)} + 5}{- \frac{39 (3.3323203500)^{2}}{100} + \cos{\left((3.3323203500) \right)}} = 3.3323203500$