Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{489 x^{3}}{500} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{489 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 9}{- \frac{1467 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{489 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 9}{- \frac{1467 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.3698027186$ $LaTeX: x_{2} = (2.3698027186) - \frac{- \frac{489 (2.3698027186)^{3}}{500} + \sin{\left((2.3698027186) \right)} + 9}{- \frac{1467 (2.3698027186)^{2}}{500} + \cos{\left((2.3698027186) \right)}} = 2.1767972404$ $LaTeX: x_{3} = (2.1767972404) - \frac{- \frac{489 (2.1767972404)^{3}}{500} + \sin{\left((2.1767972404) \right)} + 9}{- \frac{1467 (2.1767972404)^{2}}{500} + \cos{\left((2.1767972404) \right)}} = 2.1584322019$ $LaTeX: x_{4} = (2.1584322019) - \frac{- \frac{489 (2.1584322019)^{3}}{500} + \sin{\left((2.1584322019) \right)} + 9}{- \frac{1467 (2.1584322019)^{2}}{500} + \cos{\left((2.1584322019) \right)}} = 2.1582713955$ $LaTeX: x_{5} = (2.1582713955) - \frac{- \frac{489 (2.1582713955)^{3}}{500} + \sin{\left((2.1582713955) \right)} + 9}{- \frac{1467 (2.1582713955)^{2}}{500} + \cos{\left((2.1582713955) \right)}} = 2.1582713832$