Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{409 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{409 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 2}{- \frac{1227 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{409 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{1227 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 4.5422734132$ $LaTeX: x_{2} = (4.5422734132) - \frac{- \frac{409 (4.5422734132)^{3}}{1000} + \sin{\left((4.5422734132) \right)} + 2}{- \frac{1227 (4.5422734132)^{2}}{1000} + \cos{\left((4.5422734132) \right)}} = 3.0780453925$ $LaTeX: x_{3} = (3.0780453925) - \frac{- \frac{409 (3.0780453925)^{3}}{1000} + \sin{\left((3.0780453925) \right)} + 2}{- \frac{1227 (3.0780453925)^{2}}{1000} + \cos{\left((3.0780453925) \right)}} = 2.2966189044$ $LaTeX: x_{4} = (2.2966189044) - \frac{- \frac{409 (2.2966189044)^{3}}{1000} + \sin{\left((2.2966189044) \right)} + 2}{- \frac{1227 (2.2966189044)^{2}}{1000} + \cos{\left((2.2966189044) \right)}} = 1.9873999391$ $LaTeX: x_{5} = (1.9873999391) - \frac{- \frac{409 (1.9873999391)^{3}}{1000} + \sin{\left((1.9873999391) \right)} + 2}{- \frac{1227 (1.9873999391)^{2}}{1000} + \cos{\left((1.9873999391) \right)}} = 1.9310147740$