Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{793 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{793 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 9}{- \frac{2379 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{793 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{2379 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.3782053695$ $LaTeX: x_{2} = (2.3782053695) - \frac{- \frac{793 (2.3782053695)^{3}}{1000} + \cos{\left((2.3782053695) \right)} + 9}{- \frac{2379 (2.3782053695)^{2}}{1000} - \sin{\left((2.3782053695) \right)}} = 2.2093329611$ $LaTeX: x_{3} = (2.2093329611) - \frac{- \frac{793 (2.2093329611)^{3}}{1000} + \cos{\left((2.2093329611) \right)} + 9}{- \frac{2379 (2.2093329611)^{2}}{1000} - \sin{\left((2.2093329611) \right)}} = 2.1974278534$ $LaTeX: x_{4} = (2.1974278534) - \frac{- \frac{793 (2.1974278534)^{3}}{1000} + \cos{\left((2.1974278534) \right)} + 9}{- \frac{2379 (2.1974278534)^{2}}{1000} - \sin{\left((2.1974278534) \right)}} = 2.1973708016$ $LaTeX: x_{5} = (2.1973708016) - \frac{- \frac{793 (2.1973708016)^{3}}{1000} + \cos{\left((2.1973708016) \right)} + 9}{- \frac{2379 (2.1973708016)^{2}}{1000} - \sin{\left((2.1973708016) \right)}} = 2.1973708003$