Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{397 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{397 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 4}{- \frac{1191 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{397 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 4}{- \frac{1191 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.2901558647$ $LaTeX: x_{2} = (2.2901558647) - \frac{- \frac{397 (2.2901558647)^{3}}{1000} + \cos{\left((2.2901558647) \right)} + 4}{- \frac{1191 (2.2901558647)^{2}}{1000} - \sin{\left((2.2901558647) \right)}} = 2.0862001907$ $LaTeX: x_{3} = (2.0862001907) - \frac{- \frac{397 (2.0862001907)^{3}}{1000} + \cos{\left((2.0862001907) \right)} + 4}{- \frac{1191 (2.0862001907)^{2}}{1000} - \sin{\left((2.0862001907) \right)}} = 2.0700944607$ $LaTeX: x_{4} = (2.0700944607) - \frac{- \frac{397 (2.0700944607)^{3}}{1000} + \cos{\left((2.0700944607) \right)} + 4}{- \frac{1191 (2.0700944607)^{2}}{1000} - \sin{\left((2.0700944607) \right)}} = 2.0699975767$ $LaTeX: x_{5} = (2.0699975767) - \frac{- \frac{397 (2.0699975767)^{3}}{1000} + \cos{\left((2.0699975767) \right)} + 4}{- \frac{1191 (2.0699975767)^{2}}{1000} - \sin{\left((2.0699975767) \right)}} = 2.0699975732$