Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{337 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{337 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 5}{- \frac{1011 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{337 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{1011 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.4492503894$ $LaTeX: x_{2} = (2.4492503894) - \frac{- \frac{337 (2.4492503894)^{3}}{1000} + \cos{\left((2.4492503894) \right)} + 5}{- \frac{1011 (2.4492503894)^{2}}{1000} - \sin{\left((2.4492503894) \right)}} = 2.3416638442$ $LaTeX: x_{3} = (2.3416638442) - \frac{- \frac{337 (2.3416638442)^{3}}{1000} + \cos{\left((2.3416638442) \right)} + 5}{- \frac{1011 (2.3416638442)^{2}}{1000} - \sin{\left((2.3416638442) \right)}} = 2.3378427796$ $LaTeX: x_{4} = (2.3378427796) - \frac{- \frac{337 (2.3378427796)^{3}}{1000} + \cos{\left((2.3378427796) \right)} + 5}{- \frac{1011 (2.3378427796)^{2}}{1000} - \sin{\left((2.3378427796) \right)}} = 2.3378380615$ $LaTeX: x_{5} = (2.3378380615) - \frac{- \frac{337 (2.3378380615)^{3}}{1000} + \cos{\left((2.3378380615) \right)} + 5}{- \frac{1011 (2.3378380615)^{2}}{1000} - \sin{\left((2.3378380615) \right)}} = 2.3378380615$