Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{557 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{557 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 7}{- \frac{1671 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{557 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{1671 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.4052094126$ $LaTeX: x_{2} = (2.4052094126) - \frac{- \frac{557 (2.4052094126)^{3}}{1000} + \cos{\left((2.4052094126) \right)} + 7}{- \frac{1671 (2.4052094126)^{2}}{1000} - \sin{\left((2.4052094126) \right)}} = 2.2609782430$ $LaTeX: x_{3} = (2.2609782430) - \frac{- \frac{557 (2.2609782430)^{3}}{1000} + \cos{\left((2.2609782430) \right)} + 7}{- \frac{1671 (2.2609782430)^{2}}{1000} - \sin{\left((2.2609782430) \right)}} = 2.2529704404$ $LaTeX: x_{4} = (2.2529704404) - \frac{- \frac{557 (2.2529704404)^{3}}{1000} + \cos{\left((2.2529704404) \right)} + 7}{- \frac{1671 (2.2529704404)^{2}}{1000} - \sin{\left((2.2529704404) \right)}} = 2.2529465004$ $LaTeX: x_{5} = (2.2529465004) - \frac{- \frac{557 (2.2529465004)^{3}}{1000} + \cos{\left((2.2529465004) \right)} + 7}{- \frac{1671 (2.2529465004)^{2}}{1000} - \sin{\left((2.2529465004) \right)}} = 2.2529465001$