Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{39 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{39 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 8}{- \frac{117 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{39 (5.0000000000)^{3}}{500} + \sin{\left((5.0000000000) \right)} + 8}{- \frac{117 (5.0000000000)^{2}}{500} + \cos{\left((5.0000000000) \right)}} = 4.5133381471$ $LaTeX: x_{2} = (4.5133381471) - \frac{- \frac{39 (4.5133381471)^{3}}{500} + \sin{\left((4.5133381471) \right)} + 8}{- \frac{117 (4.5133381471)^{2}}{500} + \cos{\left((4.5133381471) \right)}} = 4.4828418276$ $LaTeX: x_{3} = (4.4828418276) - \frac{- \frac{39 (4.4828418276)^{3}}{500} + \sin{\left((4.4828418276) \right)} + 8}{- \frac{117 (4.4828418276)^{2}}{500} + \cos{\left((4.4828418276) \right)}} = 4.4827353064$ $LaTeX: x_{4} = (4.4827353064) - \frac{- \frac{39 (4.4827353064)^{3}}{500} + \sin{\left((4.4827353064) \right)} + 8}{- \frac{117 (4.4827353064)^{2}}{500} + \cos{\left((4.4827353064) \right)}} = 4.4827353051$ $LaTeX: x_{5} = (4.4827353051) - \frac{- \frac{39 (4.4827353051)^{3}}{500} + \sin{\left((4.4827353051) \right)} + 8}{- \frac{117 (4.4827353051)^{2}}{500} + \cos{\left((4.4827353051) \right)}} = 4.4827353051$