Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{27 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{27 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 2}{- \frac{81 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{27 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 2}{- \frac{81 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 3.2790531462$ $LaTeX: x_{2} = (3.2790531462) - \frac{- \frac{27 (3.2790531462)^{3}}{500} + \sin{\left((3.2790531462) \right)} + 2}{- \frac{81 (3.2790531462)^{2}}{500} + \cos{\left((3.2790531462) \right)}} = 3.2640825237$ $LaTeX: x_{3} = (3.2640825237) - \frac{- \frac{27 (3.2640825237)^{3}}{500} + \sin{\left((3.2640825237) \right)} + 2}{- \frac{81 (3.2640825237)^{2}}{500} + \cos{\left((3.2640825237) \right)}} = 3.2640442409$ $LaTeX: x_{4} = (3.2640442409) - \frac{- \frac{27 (3.2640442409)^{3}}{500} + \sin{\left((3.2640442409) \right)} + 2}{- \frac{81 (3.2640442409)^{2}}{500} + \cos{\left((3.2640442409) \right)}} = 3.2640442407$ $LaTeX: x_{5} = (3.2640442407) - \frac{- \frac{27 (3.2640442407)^{3}}{500} + \sin{\left((3.2640442407) \right)} + 2}{- \frac{81 (3.2640442407)^{2}}{500} + \cos{\left((3.2640442407) \right)}} = 3.2640442407$