Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{13 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=9$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{13 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 8}{- \frac{39 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 9$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (9.0000000000) - \frac{- \frac{13 (9.0000000000)^{3}}{1000} + \cos{\left((9.0000000000) \right)} + 8}{- \frac{39 (9.0000000000)^{2}}{1000} - \sin{\left((9.0000000000) \right)}} = 8.3312654644$ $LaTeX: x_{2} = (8.3312654644) - \frac{- \frac{13 (8.3312654644)^{3}}{1000} + \cos{\left((8.3312654644) \right)} + 8}{- \frac{39 (8.3312654644)^{2}}{1000} - \sin{\left((8.3312654644) \right)}} = 8.3376858214$ $LaTeX: x_{3} = (8.3376858214) - \frac{- \frac{13 (8.3376858214)^{3}}{1000} + \cos{\left((8.3376858214) \right)} + 8}{- \frac{39 (8.3376858214)^{2}}{1000} - \sin{\left((8.3376858214) \right)}} = 8.3376847398$ $LaTeX: x_{4} = (8.3376847398) - \frac{- \frac{13 (8.3376847398)^{3}}{1000} + \cos{\left((8.3376847398) \right)} + 8}{- \frac{39 (8.3376847398)^{2}}{1000} - \sin{\left((8.3376847398) \right)}} = 8.3376847398$ $LaTeX: x_{5} = (8.3376847398) - \frac{- \frac{13 (8.3376847398)^{3}}{1000} + \cos{\left((8.3376847398) \right)} + 8}{- \frac{39 (8.3376847398)^{2}}{1000} - \sin{\left((8.3376847398) \right)}} = 8.3376847398$