Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{129 x^{3}}{200} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{129 x_{n}^{3}}{200} + \sin{\left(x_{n} \right)} + 9}{- \frac{387 x_{n}^{2}}{200} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{129 (3.0000000000)^{3}}{200} + \sin{\left((3.0000000000) \right)} + 9}{- \frac{387 (3.0000000000)^{2}}{200} + \cos{\left((3.0000000000) \right)}} = 2.5504545849$ $LaTeX: x_{2} = (2.5504545849) - \frac{- \frac{129 (2.5504545849)^{3}}{200} + \sin{\left((2.5504545849) \right)} + 9}{- \frac{387 (2.5504545849)^{2}}{200} + \cos{\left((2.5504545849) \right)}} = 2.4652350763$ $LaTeX: x_{3} = (2.4652350763) - \frac{- \frac{129 (2.4652350763)^{3}}{200} + \sin{\left((2.4652350763) \right)} + 9}{- \frac{387 (2.4652350763)^{2}}{200} + \cos{\left((2.4652350763) \right)}} = 2.4622405943$ $LaTeX: x_{4} = (2.4622405943) - \frac{- \frac{129 (2.4622405943)^{3}}{200} + \sin{\left((2.4622405943) \right)} + 9}{- \frac{387 (2.4622405943)^{2}}{200} + \cos{\left((2.4622405943) \right)}} = 2.4622369516$ $LaTeX: x_{5} = (2.4622369516) - \frac{- \frac{129 (2.4622369516)^{3}}{200} + \sin{\left((2.4622369516) \right)} + 9}{- \frac{387 (2.4622369516)^{2}}{200} + \cos{\left((2.4622369516) \right)}} = 2.4622369516$