Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{193 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{193 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 8}{- \frac{579 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{193 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{579 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.4182062674$ $LaTeX: x_{2} = (2.4182062674) - \frac{- \frac{193 (2.4182062674)^{3}}{250} + \sin{\left((2.4182062674) \right)} + 8}{- \frac{579 (2.4182062674)^{2}}{250} + \cos{\left((2.4182062674) \right)}} = 2.2604409332$ $LaTeX: x_{3} = (2.2604409332) - \frac{- \frac{193 (2.2604409332)^{3}}{250} + \sin{\left((2.2604409332) \right)} + 8}{- \frac{579 (2.2604409332)^{2}}{250} + \cos{\left((2.2604409332) \right)}} = 2.2488069598$ $LaTeX: x_{4} = (2.2488069598) - \frac{- \frac{193 (2.2488069598)^{3}}{250} + \sin{\left((2.2488069598) \right)} + 8}{- \frac{579 (2.2488069598)^{2}}{250} + \cos{\left((2.2488069598) \right)}} = 2.2487453905$ $LaTeX: x_{5} = (2.2487453905) - \frac{- \frac{193 (2.2487453905)^{3}}{250} + \sin{\left((2.2487453905) \right)} + 8}{- \frac{579 (2.2487453905)^{2}}{250} + \cos{\left((2.2487453905) \right)}} = 2.2487453888$