Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{843 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{843 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{2529 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{843 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{2529 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3844518289$ $LaTeX: x_{2} = (2.3844518289) - \frac{- \frac{843 (2.3844518289)^{3}}{1000} + \sin{\left((2.3844518289) \right)} + 8}{- \frac{2529 (2.3844518289)^{2}}{1000} + \cos{\left((2.3844518289) \right)}} = 2.2029470355$ $LaTeX: x_{3} = (2.2029470355) - \frac{- \frac{843 (2.2029470355)^{3}}{1000} + \sin{\left((2.2029470355) \right)} + 8}{- \frac{2529 (2.2029470355)^{2}}{1000} + \cos{\left((2.2029470355) \right)}} = 2.1869624544$ $LaTeX: x_{4} = (2.1869624544) - \frac{- \frac{843 (2.1869624544)^{3}}{1000} + \sin{\left((2.1869624544) \right)} + 8}{- \frac{2529 (2.1869624544)^{2}}{1000} + \cos{\left((2.1869624544) \right)}} = 2.1868422428$ $LaTeX: x_{5} = (2.1868422428) - \frac{- \frac{843 (2.1868422428)^{3}}{1000} + \sin{\left((2.1868422428) \right)} + 8}{- \frac{2529 (2.1868422428)^{2}}{1000} + \cos{\left((2.1868422428) \right)}} = 2.1868422360$