Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{21 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{21 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 8}{- \frac{63 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{21 (5.0000000000)^{3}}{250} + \cos{\left((5.0000000000) \right)} + 8}{- \frac{63 (5.0000000000)^{2}}{250} - \sin{\left((5.0000000000) \right)}} = 4.5850390580$ $LaTeX: x_{2} = (4.5850390580) - \frac{- \frac{21 (4.5850390580)^{3}}{250} + \cos{\left((4.5850390580) \right)} + 8}{- \frac{63 (4.5850390580)^{2}}{250} - \sin{\left((4.5850390580) \right)}} = 4.5330828135$ $LaTeX: x_{3} = (4.5330828135) - \frac{- \frac{21 (4.5330828135)^{3}}{250} + \cos{\left((4.5330828135) \right)} + 8}{- \frac{63 (4.5330828135)^{2}}{250} - \sin{\left((4.5330828135) \right)}} = 4.5323883830$ $LaTeX: x_{4} = (4.5323883830) - \frac{- \frac{21 (4.5323883830)^{3}}{250} + \cos{\left((4.5323883830) \right)} + 8}{- \frac{63 (4.5323883830)^{2}}{250} - \sin{\left((4.5323883830) \right)}} = 4.5323882619$ $LaTeX: x_{5} = (4.5323882619) - \frac{- \frac{21 (4.5323882619)^{3}}{250} + \cos{\left((4.5323882619) \right)} + 8}{- \frac{63 (4.5323882619)^{2}}{250} - \sin{\left((4.5323882619) \right)}} = 4.5323882619$