Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{9 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{9 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 6}{- \frac{27 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{9 (5.0000000000)^{3}}{250} + \cos{\left((5.0000000000) \right)} + 6}{- \frac{27 (5.0000000000)^{2}}{250} - \sin{\left((5.0000000000) \right)}} = 6.0244598552$ $LaTeX: x_{2} = (6.0244598552) - \frac{- \frac{9 (6.0244598552)^{3}}{250} + \cos{\left((6.0244598552) \right)} + 6}{- \frac{27 (6.0244598552)^{2}}{250} - \sin{\left((6.0244598552) \right)}} = 5.7775187984$ $LaTeX: x_{3} = (5.7775187984) - \frac{- \frac{9 (5.7775187984)^{3}}{250} + \cos{\left((5.7775187984) \right)} + 6}{- \frac{27 (5.7775187984)^{2}}{250} - \sin{\left((5.7775187984) \right)}} = 5.7557861087$ $LaTeX: x_{4} = (5.7557861087) - \frac{- \frac{9 (5.7557861087)^{3}}{250} + \cos{\left((5.7557861087) \right)} + 6}{- \frac{27 (5.7557861087)^{2}}{250} - \sin{\left((5.7557861087) \right)}} = 5.7556234554$ $LaTeX: x_{5} = (5.7556234554) - \frac{- \frac{9 (5.7556234554)^{3}}{250} + \cos{\left((5.7556234554) \right)} + 6}{- \frac{27 (5.7556234554)^{2}}{250} - \sin{\left((5.7556234554) \right)}} = 5.7556234464$