Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{247 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{247 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 4}{- \frac{741 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{247 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 4}{- \frac{741 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.4627124790$ $LaTeX: x_{2} = (2.4627124790) - \frac{- \frac{247 (2.4627124790)^{3}}{1000} + \cos{\left((2.4627124790) \right)} + 4}{- \frac{741 (2.4627124790)^{2}}{1000} - \sin{\left((2.4627124790) \right)}} = 2.3714353142$ $LaTeX: x_{3} = (2.3714353142) - \frac{- \frac{247 (2.3714353142)^{3}}{1000} + \cos{\left((2.3714353142) \right)} + 4}{- \frac{741 (2.3714353142)^{2}}{1000} - \sin{\left((2.3714353142) \right)}} = 2.3689975626$ $LaTeX: x_{4} = (2.3689975626) - \frac{- \frac{247 (2.3689975626)^{3}}{1000} + \cos{\left((2.3689975626) \right)} + 4}{- \frac{741 (2.3689975626)^{2}}{1000} - \sin{\left((2.3689975626) \right)}} = 2.3689958520$ $LaTeX: x_{5} = (2.3689958520) - \frac{- \frac{247 (2.3689958520)^{3}}{1000} + \cos{\left((2.3689958520) \right)} + 4}{- \frac{741 (2.3689958520)^{2}}{1000} - \sin{\left((2.3689958520) \right)}} = 2.3689958520$