Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{949 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{949 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 9}{- \frac{2847 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{949 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 9}{- \frac{2847 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3806829505$ $LaTeX: x_{2} = (2.3806829505) - \frac{- \frac{949 (2.3806829505)^{3}}{1000} + \sin{\left((2.3806829505) \right)} + 9}{- \frac{2847 (2.3806829505)^{2}}{1000} + \cos{\left((2.3806829505) \right)}} = 2.1959165740$ $LaTeX: x_{3} = (2.1959165740) - \frac{- \frac{949 (2.1959165740)^{3}}{1000} + \sin{\left((2.1959165740) \right)} + 9}{- \frac{2847 (2.1959165740)^{2}}{1000} + \cos{\left((2.1959165740) \right)}} = 2.1792962529$ $LaTeX: x_{4} = (2.1792962529) - \frac{- \frac{949 (2.1792962529)^{3}}{1000} + \sin{\left((2.1792962529) \right)} + 9}{- \frac{2847 (2.1792962529)^{2}}{1000} + \cos{\left((2.1792962529) \right)}} = 2.1791660430$ $LaTeX: x_{5} = (2.1791660430) - \frac{- \frac{949 (2.1791660430)^{3}}{1000} + \sin{\left((2.1791660430) \right)} + 9}{- \frac{2847 (2.1791660430)^{2}}{1000} + \cos{\left((2.1791660430) \right)}} = 2.1791660351$