Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{83 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{83 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 8}{- \frac{249 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{83 (5.0000000000)^{3}}{1000} + \cos{\left((5.0000000000) \right)} + 8}{- \frac{249 (5.0000000000)^{2}}{1000} - \sin{\left((5.0000000000) \right)}} = 4.6028659815$ $LaTeX: x_{2} = (4.6028659815) - \frac{- \frac{83 (4.6028659815)^{3}}{1000} + \cos{\left((4.6028659815) \right)} + 8}{- \frac{249 (4.6028659815)^{2}}{1000} - \sin{\left((4.6028659815) \right)}} = 4.5553810416$ $LaTeX: x_{3} = (4.5553810416) - \frac{- \frac{83 (4.5553810416)^{3}}{1000} + \cos{\left((4.5553810416) \right)} + 8}{- \frac{249 (4.5553810416)^{2}}{1000} - \sin{\left((4.5553810416) \right)}} = 4.5547985571$ $LaTeX: x_{4} = (4.5547985571) - \frac{- \frac{83 (4.5547985571)^{3}}{1000} + \cos{\left((4.5547985571) \right)} + 8}{- \frac{249 (4.5547985571)^{2}}{1000} - \sin{\left((4.5547985571) \right)}} = 4.5547984714$ $LaTeX: x_{5} = (4.5547984714) - \frac{- \frac{83 (4.5547984714)^{3}}{1000} + \cos{\left((4.5547984714) \right)} + 8}{- \frac{249 (4.5547984714)^{2}}{1000} - \sin{\left((4.5547984714) \right)}} = 4.5547984714$