Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{639 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{639 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 7}{- \frac{1917 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{639 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{1917 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.3536325786$ $LaTeX: x_{2} = (2.3536325786) - \frac{- \frac{639 (2.3536325786)^{3}}{1000} + \cos{\left((2.3536325786) \right)} + 7}{- \frac{1917 (2.3536325786)^{2}}{1000} - \sin{\left((2.3536325786) \right)}} = 2.1738464889$ $LaTeX: x_{3} = (2.1738464889) - \frac{- \frac{639 (2.1738464889)^{3}}{1000} + \cos{\left((2.1738464889) \right)} + 7}{- \frac{1917 (2.1738464889)^{2}}{1000} - \sin{\left((2.1738464889) \right)}} = 2.1605460485$ $LaTeX: x_{4} = (2.1605460485) - \frac{- \frac{639 (2.1605460485)^{3}}{1000} + \cos{\left((2.1605460485) \right)} + 7}{- \frac{1917 (2.1605460485)^{2}}{1000} - \sin{\left((2.1605460485) \right)}} = 2.1604759174$ $LaTeX: x_{5} = (2.1604759174) - \frac{- \frac{639 (2.1604759174)^{3}}{1000} + \cos{\left((2.1604759174) \right)} + 7}{- \frac{1917 (2.1604759174)^{2}}{1000} - \sin{\left((2.1604759174) \right)}} = 2.1604759155$