Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{613 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{613 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 8}{- \frac{1839 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{613 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 8}{- \frac{1839 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.4284133776$ $LaTeX: x_{2} = (2.4284133776) - \frac{- \frac{613 (2.4284133776)^{3}}{1000} + \cos{\left((2.4284133776) \right)} + 8}{- \frac{1839 (2.4284133776)^{2}}{1000} - \sin{\left((2.4284133776) \right)}} = 2.2949301161$ $LaTeX: x_{3} = (2.2949301161) - \frac{- \frac{613 (2.2949301161)^{3}}{1000} + \cos{\left((2.2949301161) \right)} + 8}{- \frac{1839 (2.2949301161)^{2}}{1000} - \sin{\left((2.2949301161) \right)}} = 2.2880639721$ $LaTeX: x_{4} = (2.2880639721) - \frac{- \frac{613 (2.2880639721)^{3}}{1000} + \cos{\left((2.2880639721) \right)} + 8}{- \frac{1839 (2.2880639721)^{2}}{1000} - \sin{\left((2.2880639721) \right)}} = 2.2880463257$ $LaTeX: x_{5} = (2.2880463257) - \frac{- \frac{613 (2.2880463257)^{3}}{1000} + \cos{\left((2.2880463257) \right)} + 8}{- \frac{1839 (2.2880463257)^{2}}{1000} - \sin{\left((2.2880463257) \right)}} = 2.2880463255$