Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{249 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{249 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 8}{- \frac{747 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{249 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{747 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.6325240545$ $LaTeX: x_{2} = (2.6325240545) - \frac{- \frac{249 (2.6325240545)^{3}}{500} + \sin{\left((2.6325240545) \right)} + 8}{- \frac{747 (2.6325240545)^{2}}{500} + \cos{\left((2.6325240545) \right)}} = 2.5792515478$ $LaTeX: x_{3} = (2.5792515478) - \frac{- \frac{249 (2.5792515478)^{3}}{500} + \sin{\left((2.5792515478) \right)} + 8}{- \frac{747 (2.5792515478)^{2}}{500} + \cos{\left((2.5792515478) \right)}} = 2.5781574463$ $LaTeX: x_{4} = (2.5781574463) - \frac{- \frac{249 (2.5781574463)^{3}}{500} + \sin{\left((2.5781574463) \right)} + 8}{- \frac{747 (2.5781574463)^{2}}{500} + \cos{\left((2.5781574463) \right)}} = 2.5781569887$ $LaTeX: x_{5} = (2.5781569887) - \frac{- \frac{249 (2.5781569887)^{3}}{500} + \sin{\left((2.5781569887) \right)} + 8}{- \frac{747 (2.5781569887)^{2}}{500} + \cos{\left((2.5781569887) \right)}} = 2.5781569887$