Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{111 x^{3}}{250} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{111 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 5}{- \frac{333 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{111 (3.0000000000)^{3}}{250} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{333 (3.0000000000)^{2}}{250} - \sin{\left((3.0000000000) \right)}} = 2.3422447390$ $LaTeX: x_{2} = (2.3422447390) - \frac{- \frac{111 (2.3422447390)^{3}}{250} + \cos{\left((2.3422447390) \right)} + 5}{- \frac{333 (2.3422447390)^{2}}{250} - \sin{\left((2.3422447390) \right)}} = 2.1674663999$ $LaTeX: x_{3} = (2.1674663999) - \frac{- \frac{111 (2.1674663999)^{3}}{250} + \cos{\left((2.1674663999) \right)} + 5}{- \frac{333 (2.1674663999)^{2}}{250} - \sin{\left((2.1674663999) \right)}} = 2.1557583990$ $LaTeX: x_{4} = (2.1557583990) - \frac{- \frac{111 (2.1557583990)^{3}}{250} + \cos{\left((2.1557583990) \right)} + 5}{- \frac{333 (2.1557583990)^{2}}{250} - \sin{\left((2.1557583990) \right)}} = 2.1557076085$ $LaTeX: x_{5} = (2.1557076085) - \frac{- \frac{111 (2.1557076085)^{3}}{250} + \cos{\left((2.1557076085) \right)} + 5}{- \frac{333 (2.1557076085)^{2}}{250} - \sin{\left((2.1557076085) \right)}} = 2.1557076076$