Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{409 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{409 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 4}{- \frac{1227 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{409 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 4}{- \frac{1227 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.2817501519$ $LaTeX: x_{2} = (2.2817501519) - \frac{- \frac{409 (2.2817501519)^{3}}{1000} + \cos{\left((2.2817501519) \right)} + 4}{- \frac{1227 (2.2817501519)^{2}}{1000} - \sin{\left((2.2817501519) \right)}} = 2.0702547177$ $LaTeX: x_{3} = (2.0702547177) - \frac{- \frac{409 (2.0702547177)^{3}}{1000} + \cos{\left((2.0702547177) \right)} + 4}{- \frac{1227 (2.0702547177)^{2}}{1000} - \sin{\left((2.0702547177) \right)}} = 2.0526533582$ $LaTeX: x_{4} = (2.0526533582) - \frac{- \frac{409 (2.0526533582)^{3}}{1000} + \cos{\left((2.0526533582) \right)} + 4}{- \frac{1227 (2.0526533582)^{2}}{1000} - \sin{\left((2.0526533582) \right)}} = 2.0525358950$ $LaTeX: x_{5} = (2.0525358950) - \frac{- \frac{409 (2.0525358950)^{3}}{1000} + \cos{\left((2.0525358950) \right)} + 4}{- \frac{1227 (2.0525358950)^{2}}{1000} - \sin{\left((2.0525358950) \right)}} = 2.0525358898$