Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{249 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{249 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{747 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{249 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{747 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.4247101476$ $LaTeX: x_{2} = (2.4247101476) - \frac{- \frac{249 (2.4247101476)^{3}}{500} + \sin{\left((2.4247101476) \right)} + 5}{- \frac{747 (2.4247101476)^{2}}{500} + \cos{\left((2.4247101476) \right)}} = 2.2735000581$ $LaTeX: x_{3} = (2.2735000581) - \frac{- \frac{249 (2.2735000581)^{3}}{500} + \sin{\left((2.2735000581) \right)} + 5}{- \frac{747 (2.2735000581)^{2}}{500} + \cos{\left((2.2735000581) \right)}} = 2.2628605726$ $LaTeX: x_{4} = (2.2628605726) - \frac{- \frac{249 (2.2628605726)^{3}}{500} + \sin{\left((2.2628605726) \right)} + 5}{- \frac{747 (2.2628605726)^{2}}{500} + \cos{\left((2.2628605726) \right)}} = 2.2628090281$ $LaTeX: x_{5} = (2.2628090281) - \frac{- \frac{249 (2.2628090281)^{3}}{500} + \sin{\left((2.2628090281) \right)} + 5}{- \frac{747 (2.2628090281)^{2}}{500} + \cos{\left((2.2628090281) \right)}} = 2.2628090269$