Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{813 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{813 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{2439 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{813 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{2439 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3980260447$ $LaTeX: x_{2} = (2.3980260447) - \frac{- \frac{813 (2.3980260447)^{3}}{1000} + \sin{\left((2.3980260447) \right)} + 8}{- \frac{2439 (2.3980260447)^{2}}{1000} + \cos{\left((2.3980260447) \right)}} = 2.2263450516$ $LaTeX: x_{3} = (2.2263450516) - \frac{- \frac{813 (2.2263450516)^{3}}{1000} + \sin{\left((2.2263450516) \right)} + 8}{- \frac{2439 (2.2263450516)^{2}}{1000} + \cos{\left((2.2263450516) \right)}} = 2.2122612859$ $LaTeX: x_{4} = (2.2122612859) - \frac{- \frac{813 (2.2122612859)^{3}}{1000} + \sin{\left((2.2122612859) \right)} + 8}{- \frac{2439 (2.2122612859)^{2}}{1000} + \cos{\left((2.2122612859) \right)}} = 2.2121692486$ $LaTeX: x_{5} = (2.2121692486) - \frac{- \frac{813 (2.2121692486)^{3}}{1000} + \sin{\left((2.2121692486) \right)} + 8}{- \frac{2439 (2.2121692486)^{2}}{1000} + \cos{\left((2.2121692486) \right)}} = 2.2121692447$