Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{11 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=9$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{11 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 5}{- \frac{33 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 9$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (9.0000000000) - \frac{- \frac{11 (9.0000000000)^{3}}{1000} + \sin{\left((9.0000000000) \right)} + 5}{- \frac{33 (9.0000000000)^{2}}{1000} + \cos{\left((9.0000000000) \right)}} = 8.2726599414$ $LaTeX: x_{2} = (8.2726599414) - \frac{- \frac{11 (8.2726599414)^{3}}{1000} + \sin{\left((8.2726599414) \right)} + 5}{- \frac{33 (8.2726599414)^{2}}{1000} + \cos{\left((8.2726599414) \right)}} = 8.1548048663$ $LaTeX: x_{3} = (8.1548048663) - \frac{- \frac{11 (8.1548048663)^{3}}{1000} + \sin{\left((8.1548048663) \right)} + 5}{- \frac{33 (8.1548048663)^{2}}{1000} + \cos{\left((8.1548048663) \right)}} = 8.1507008430$ $LaTeX: x_{4} = (8.1507008430) - \frac{- \frac{11 (8.1507008430)^{3}}{1000} + \sin{\left((8.1507008430) \right)} + 5}{- \frac{33 (8.1507008430)^{2}}{1000} + \cos{\left((8.1507008430) \right)}} = 8.1506957806$ $LaTeX: x_{5} = (8.1506957806) - \frac{- \frac{11 (8.1506957806)^{3}}{1000} + \sin{\left((8.1506957806) \right)} + 5}{- \frac{33 (8.1506957806)^{2}}{1000} + \cos{\left((8.1506957806) \right)}} = 8.1506957805$