Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{14 x^{3}}{125} - 9$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{14 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 9}{- \frac{42 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{14 (5.0000000000)^{3}}{125} + \sin{\left((5.0000000000) \right)} + 9}{- \frac{42 (5.0000000000)^{2}}{125} + \cos{\left((5.0000000000) \right)}} = 4.2658112056$ $LaTeX: x_{2} = (4.2658112056) - \frac{- \frac{14 (4.2658112056)^{3}}{125} + \sin{\left((4.2658112056) \right)} + 9}{- \frac{42 (4.2658112056)^{2}}{125} + \cos{\left((4.2658112056) \right)}} = 4.1747655462$ $LaTeX: x_{3} = (4.1747655462) - \frac{- \frac{14 (4.1747655462)^{3}}{125} + \sin{\left((4.1747655462) \right)} + 9}{- \frac{42 (4.1747655462)^{2}}{125} + \cos{\left((4.1747655462) \right)}} = 4.1734911781$ $LaTeX: x_{4} = (4.1734911781) - \frac{- \frac{14 (4.1734911781)^{3}}{125} + \sin{\left((4.1734911781) \right)} + 9}{- \frac{42 (4.1734911781)^{2}}{125} + \cos{\left((4.1734911781) \right)}} = 4.1734909298$ $LaTeX: x_{5} = (4.1734909298) - \frac{- \frac{14 (4.1734909298)^{3}}{125} + \sin{\left((4.1734909298) \right)} + 9}{- \frac{42 (4.1734909298)^{2}}{125} + \cos{\left((4.1734909298) \right)}} = 4.1734909298$