Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{407 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{407 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 8}{- \frac{1221 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{407 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 8}{- \frac{1221 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.6425022827$ $LaTeX: x_{2} = (2.6425022827) - \frac{- \frac{407 (2.6425022827)^{3}}{1000} + \cos{\left((2.6425022827) \right)} + 8}{- \frac{1221 (2.6425022827)^{2}}{1000} - \sin{\left((2.6425022827) \right)}} = 2.5994102689$ $LaTeX: x_{3} = (2.5994102689) - \frac{- \frac{407 (2.5994102689)^{3}}{1000} + \cos{\left((2.5994102689) \right)} + 8}{- \frac{1221 (2.5994102689)^{2}}{1000} - \sin{\left((2.5994102689) \right)}} = 2.5988227771$ $LaTeX: x_{4} = (2.5988227771) - \frac{- \frac{407 (2.5988227771)^{3}}{1000} + \cos{\left((2.5988227771) \right)} + 8}{- \frac{1221 (2.5988227771)^{2}}{1000} - \sin{\left((2.5988227771) \right)}} = 2.5988226689$ $LaTeX: x_{5} = (2.5988226689) - \frac{- \frac{407 (2.5988226689)^{3}}{1000} + \cos{\left((2.5988226689) \right)} + 8}{- \frac{1221 (2.5988226689)^{2}}{1000} - \sin{\left((2.5988226689) \right)}} = 2.5988226689$