Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{883 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{883 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 9}{- \frac{2649 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{883 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 9}{- \frac{2649 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4080027210$ $LaTeX: x_{2} = (2.4080027210) - \frac{- \frac{883 (2.4080027210)^{3}}{1000} + \sin{\left((2.4080027210) \right)} + 9}{- \frac{2649 (2.4080027210)^{2}}{1000} + \cos{\left((2.4080027210) \right)}} = 2.2428424237$ $LaTeX: x_{3} = (2.2428424237) - \frac{- \frac{883 (2.2428424237)^{3}}{1000} + \sin{\left((2.2428424237) \right)} + 9}{- \frac{2649 (2.2428424237)^{2}}{1000} + \cos{\left((2.2428424237) \right)}} = 2.2299595157$ $LaTeX: x_{4} = (2.2299595157) - \frac{- \frac{883 (2.2299595157)^{3}}{1000} + \sin{\left((2.2299595157) \right)} + 9}{- \frac{2649 (2.2299595157)^{2}}{1000} + \cos{\left((2.2299595157) \right)}} = 2.2298833945$ $LaTeX: x_{5} = (2.2298833945) - \frac{- \frac{883 (2.2298833945)^{3}}{1000} + \sin{\left((2.2298833945) \right)} + 9}{- \frac{2649 (2.2298833945)^{2}}{1000} + \cos{\left((2.2298833945) \right)}} = 2.2298833919$