Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{7 x^{3}}{125} - 9$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{7 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 9}{- \frac{21 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{7 (5.0000000000)^{3}}{125} + \sin{\left((5.0000000000) \right)} + 9}{- \frac{21 (5.0000000000)^{2}}{125} + \cos{\left((5.0000000000) \right)}} = 5.2658288878$ $LaTeX: x_{2} = (5.2658288878) - \frac{- \frac{7 (5.2658288878)^{3}}{125} + \sin{\left((5.2658288878) \right)} + 9}{- \frac{21 (5.2658288878)^{2}}{125} + \cos{\left((5.2658288878) \right)}} = 5.2591474505$ $LaTeX: x_{3} = (5.2591474505) - \frac{- \frac{7 (5.2591474505)^{3}}{125} + \sin{\left((5.2591474505) \right)} + 9}{- \frac{21 (5.2591474505)^{2}}{125} + \cos{\left((5.2591474505) \right)}} = 5.2591424924$ $LaTeX: x_{4} = (5.2591424924) - \frac{- \frac{7 (5.2591424924)^{3}}{125} + \sin{\left((5.2591424924) \right)} + 9}{- \frac{21 (5.2591424924)^{2}}{125} + \cos{\left((5.2591424924) \right)}} = 5.2591424924$ $LaTeX: x_{5} = (5.2591424924) - \frac{- \frac{7 (5.2591424924)^{3}}{125} + \sin{\left((5.2591424924) \right)} + 9}{- \frac{21 (5.2591424924)^{2}}{125} + \cos{\left((5.2591424924) \right)}} = 5.2591424924$