Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{387 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{387 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 2}{- \frac{1161 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{387 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{1161 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.0753226000$ $LaTeX: x_{2} = (2.0753226000) - \frac{- \frac{387 (2.0753226000)^{3}}{1000} + \cos{\left((2.0753226000) \right)} + 2}{- \frac{1161 (2.0753226000)^{2}}{1000} - \sin{\left((2.0753226000) \right)}} = 1.7447235439$ $LaTeX: x_{3} = (1.7447235439) - \frac{- \frac{387 (1.7447235439)^{3}}{1000} + \cos{\left((1.7447235439) \right)} + 2}{- \frac{1161 (1.7447235439)^{2}}{1000} - \sin{\left((1.7447235439) \right)}} = 1.6941764908$ $LaTeX: x_{4} = (1.6941764908) - \frac{- \frac{387 (1.6941764908)^{3}}{1000} + \cos{\left((1.6941764908) \right)} + 2}{- \frac{1161 (1.6941764908)^{2}}{1000} - \sin{\left((1.6941764908) \right)}} = 1.6930375393$ $LaTeX: x_{5} = (1.6930375393) - \frac{- \frac{387 (1.6930375393)^{3}}{1000} + \cos{\left((1.6930375393) \right)} + 2}{- \frac{1161 (1.6930375393)^{2}}{1000} - \sin{\left((1.6930375393) \right)}} = 1.6930369673$