Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{829 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{829 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 7}{- \frac{2487 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{829 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{2487 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3478849576$ $LaTeX: x_{2} = (2.3478849576) - \frac{- \frac{829 (2.3478849576)^{3}}{1000} + \sin{\left((2.3478849576) \right)} + 7}{- \frac{2487 (2.3478849576)^{2}}{1000} + \cos{\left((2.3478849576) \right)}} = 2.1385526836$ $LaTeX: x_{3} = (2.1385526836) - \frac{- \frac{829 (2.1385526836)^{3}}{1000} + \sin{\left((2.1385526836) \right)} + 7}{- \frac{2487 (2.1385526836)^{2}}{1000} + \cos{\left((2.1385526836) \right)}} = 2.1163140276$ $LaTeX: x_{4} = (2.1163140276) - \frac{- \frac{829 (2.1163140276)^{3}}{1000} + \sin{\left((2.1163140276) \right)} + 7}{- \frac{2487 (2.1163140276)^{2}}{1000} + \cos{\left((2.1163140276) \right)}} = 2.1160712084$ $LaTeX: x_{5} = (2.1160712084) - \frac{- \frac{829 (2.1160712084)^{3}}{1000} + \sin{\left((2.1160712084) \right)} + 7}{- \frac{2487 (2.1160712084)^{2}}{1000} + \cos{\left((2.1160712084) \right)}} = 2.1160711797$