Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{907 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{907 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 9}{- \frac{2721 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{907 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{2721 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.3309414452$ $LaTeX: x_{2} = (2.3309414452) - \frac{- \frac{907 (2.3309414452)^{3}}{1000} + \cos{\left((2.3309414452) \right)} + 9}{- \frac{2721 (2.3309414452)^{2}}{1000} - \sin{\left((2.3309414452) \right)}} = 2.1261605571$ $LaTeX: x_{3} = (2.1261605571) - \frac{- \frac{907 (2.1261605571)^{3}}{1000} + \cos{\left((2.1261605571) \right)} + 9}{- \frac{2721 (2.1261605571)^{2}}{1000} - \sin{\left((2.1261605571) \right)}} = 2.1075430390$ $LaTeX: x_{4} = (2.1075430390) - \frac{- \frac{907 (2.1075430390)^{3}}{1000} + \cos{\left((2.1075430390) \right)} + 9}{- \frac{2721 (2.1075430390)^{2}}{1000} - \sin{\left((2.1075430390) \right)}} = 2.1073955778$ $LaTeX: x_{5} = (2.1073955778) - \frac{- \frac{907 (2.1073955778)^{3}}{1000} + \cos{\left((2.1073955778) \right)} + 9}{- \frac{2721 (2.1073955778)^{2}}{1000} - \sin{\left((2.1073955778) \right)}} = 2.1073955686$