Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{217 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{217 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 2}{- \frac{651 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{217 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{651 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 4.1606646618$ $LaTeX: x_{2} = (4.1606646618) - \frac{- \frac{217 (4.1606646618)^{3}}{500} + \sin{\left((4.1606646618) \right)} + 2}{- \frac{651 (4.1606646618)^{2}}{500} + \cos{\left((4.1606646618) \right)}} = 2.8550886849$ $LaTeX: x_{3} = (2.8550886849) - \frac{- \frac{217 (2.8550886849)^{3}}{500} + \sin{\left((2.8550886849) \right)} + 2}{- \frac{651 (2.8550886849)^{2}}{500} + \cos{\left((2.8550886849) \right)}} = 2.1795207456$ $LaTeX: x_{4} = (2.1795207456) - \frac{- \frac{217 (2.1795207456)^{3}}{500} + \sin{\left((2.1795207456) \right)} + 2}{- \frac{651 (2.1795207456)^{2}}{500} + \cos{\left((2.1795207456) \right)}} = 1.9319159426$ $LaTeX: x_{5} = (1.9319159426) - \frac{- \frac{217 (1.9319159426)^{3}}{500} + \sin{\left((1.9319159426) \right)} + 2}{- \frac{651 (1.9319159426)^{2}}{500} + \cos{\left((1.9319159426) \right)}} = 1.8947284368$