Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{717 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{717 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 6}{- \frac{2151 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{717 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{2151 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3504405688$ $LaTeX: x_{2} = (2.3504405688) - \frac{- \frac{717 (2.3504405688)^{3}}{1000} + \sin{\left((2.3504405688) \right)} + 6}{- \frac{2151 (2.3504405688)^{2}}{1000} + \cos{\left((2.3504405688) \right)}} = 2.1439310348$ $LaTeX: x_{3} = (2.1439310348) - \frac{- \frac{717 (2.1439310348)^{3}}{1000} + \sin{\left((2.1439310348) \right)} + 6}{- \frac{2151 (2.1439310348)^{2}}{1000} + \cos{\left((2.1439310348) \right)}} = 2.1223152020$ $LaTeX: x_{4} = (2.1223152020) - \frac{- \frac{717 (2.1223152020)^{3}}{1000} + \sin{\left((2.1223152020) \right)} + 6}{- \frac{2151 (2.1223152020)^{2}}{1000} + \cos{\left((2.1223152020) \right)}} = 2.1220856127$ $LaTeX: x_{5} = (2.1220856127) - \frac{- \frac{717 (2.1220856127)^{3}}{1000} + \sin{\left((2.1220856127) \right)} + 6}{- \frac{2151 (2.1220856127)^{2}}{1000} + \cos{\left((2.1220856127) \right)}} = 2.1220855869$