Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{407 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{407 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 5}{- \frac{1221 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{407 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 5}{- \frac{1221 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 2.4394225890$ $LaTeX: x_{2} = (2.4394225890) - \frac{- \frac{407 (2.4394225890)^{3}}{500} + \cos{\left((2.4394225890) \right)} + 5}{- \frac{1221 (2.4394225890)^{2}}{500} - \sin{\left((2.4394225890) \right)}} = 1.9400150194$ $LaTeX: x_{3} = (1.9400150194) - \frac{- \frac{407 (1.9400150194)^{3}}{500} + \cos{\left((1.9400150194) \right)} + 5}{- \frac{1221 (1.9400150194)^{2}}{500} - \sin{\left((1.9400150194) \right)}} = 1.8111706220$ $LaTeX: x_{4} = (1.8111706220) - \frac{- \frac{407 (1.8111706220)^{3}}{500} + \cos{\left((1.8111706220) \right)} + 5}{- \frac{1221 (1.8111706220)^{2}}{500} - \sin{\left((1.8111706220) \right)}} = 1.8029043104$ $LaTeX: x_{5} = (1.8029043104) - \frac{- \frac{407 (1.8029043104)^{3}}{500} + \cos{\left((1.8029043104) \right)} + 5}{- \frac{1221 (1.8029043104)^{2}}{500} - \sin{\left((1.8029043104) \right)}} = 1.8028713481$