Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{9 x^{3}}{200} - 9$ using $LaTeX: \displaystyle x_0=7$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{9 x_{n}^{3}}{200} + \cos{\left(x_{n} \right)} + 9}{- \frac{27 x_{n}^{2}}{200} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 7$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (7.0000000000) - \frac{- \frac{9 (7.0000000000)^{3}}{200} + \cos{\left((7.0000000000) \right)} + 9}{- \frac{27 (7.0000000000)^{2}}{200} - \sin{\left((7.0000000000) \right)}} = 6.2187694974$ $LaTeX: x_{2} = (6.2187694974) - \frac{- \frac{9 (6.2187694974)^{3}}{200} + \cos{\left((6.2187694974) \right)} + 9}{- \frac{27 (6.2187694974)^{2}}{200} - \sin{\left((6.2187694974) \right)}} = 6.0588679848$ $LaTeX: x_{3} = (6.0588679848) - \frac{- \frac{9 (6.0588679848)^{3}}{200} + \cos{\left((6.0588679848) \right)} + 9}{- \frac{27 (6.0588679848)^{2}}{200} - \sin{\left((6.0588679848) \right)}} = 6.0516916523$ $LaTeX: x_{4} = (6.0516916523) - \frac{- \frac{9 (6.0516916523)^{3}}{200} + \cos{\left((6.0516916523) \right)} + 9}{- \frac{27 (6.0516916523)^{2}}{200} - \sin{\left((6.0516916523) \right)}} = 6.0516773993$ $LaTeX: x_{5} = (6.0516773993) - \frac{- \frac{9 (6.0516773993)^{3}}{200} + \cos{\left((6.0516773993) \right)} + 9}{- \frac{27 (6.0516773993)^{2}}{200} - \sin{\left((6.0516773993) \right)}} = 6.0516773992$