Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{51 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{51 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 8}{- \frac{153 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{51 (5.0000000000)^{3}}{500} + \cos{\left((5.0000000000) \right)} + 8}{- \frac{153 (5.0000000000)^{2}}{500} - \sin{\left((5.0000000000) \right)}} = 4.3324933093$ $LaTeX: x_{2} = (4.3324933093) - \frac{- \frac{51 (4.3324933093)^{3}}{500} + \cos{\left((4.3324933093) \right)} + 8}{- \frac{153 (4.3324933093)^{2}}{500} - \sin{\left((4.3324933093) \right)}} = 4.1942241596$ $LaTeX: x_{3} = (4.1942241596) - \frac{- \frac{51 (4.1942241596)^{3}}{500} + \cos{\left((4.1942241596) \right)} + 8}{- \frac{153 (4.1942241596)^{2}}{500} - \sin{\left((4.1942241596) \right)}} = 4.1895437859$ $LaTeX: x_{4} = (4.1895437859) - \frac{- \frac{51 (4.1895437859)^{3}}{500} + \cos{\left((4.1895437859) \right)} + 8}{- \frac{153 (4.1895437859)^{2}}{500} - \sin{\left((4.1895437859) \right)}} = 4.1895387545$ $LaTeX: x_{5} = (4.1895387545) - \frac{- \frac{51 (4.1895387545)^{3}}{500} + \cos{\left((4.1895387545) \right)} + 8}{- \frac{153 (4.1895387545)^{2}}{500} - \sin{\left((4.1895387545) \right)}} = 4.1895387545$