Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{227 x^{3}}{500} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{227 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 9}{- \frac{681 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{227 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 9}{- \frac{681 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.7647281282$ $LaTeX: x_{2} = (2.7647281282) - \frac{- \frac{227 (2.7647281282)^{3}}{500} + \sin{\left((2.7647281282) \right)} + 9}{- \frac{681 (2.7647281282)^{2}}{500} + \cos{\left((2.7647281282) \right)}} = 2.7447741186$ $LaTeX: x_{3} = (2.7447741186) - \frac{- \frac{227 (2.7447741186)^{3}}{500} + \sin{\left((2.7447741186) \right)} + 9}{- \frac{681 (2.7447741186)^{2}}{500} + \cos{\left((2.7447741186) \right)}} = 2.7446337139$ $LaTeX: x_{4} = (2.7446337139) - \frac{- \frac{227 (2.7446337139)^{3}}{500} + \sin{\left((2.7446337139) \right)} + 9}{- \frac{681 (2.7446337139)^{2}}{500} + \cos{\left((2.7446337139) \right)}} = 2.7446337070$ $LaTeX: x_{5} = (2.7446337070) - \frac{- \frac{227 (2.7446337070)^{3}}{500} + \sin{\left((2.7446337070) \right)} + 9}{- \frac{681 (2.7446337070)^{2}}{500} + \cos{\left((2.7446337070) \right)}} = 2.7446337070$