Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{621 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{621 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 3}{- \frac{1863 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{621 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 3}{- \frac{1863 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.0794356169$ $LaTeX: x_{2} = (2.0794356169) - \frac{- \frac{621 (2.0794356169)^{3}}{1000} + \cos{\left((2.0794356169) \right)} + 3}{- \frac{1863 (2.0794356169)^{2}}{1000} - \sin{\left((2.0794356169) \right)}} = 1.7355308839$ $LaTeX: x_{3} = (1.7355308839) - \frac{- \frac{621 (1.7355308839)^{3}}{1000} + \cos{\left((1.7355308839) \right)} + 3}{- \frac{1863 (1.7355308839)^{2}}{1000} - \sin{\left((1.7355308839) \right)}} = 1.6733462972$ $LaTeX: x_{4} = (1.6733462972) - \frac{- \frac{621 (1.6733462972)^{3}}{1000} + \cos{\left((1.6733462972) \right)} + 3}{- \frac{1863 (1.6733462972)^{2}}{1000} - \sin{\left((1.6733462972) \right)}} = 1.6714020808$ $LaTeX: x_{5} = (1.6714020808) - \frac{- \frac{621 (1.6714020808)^{3}}{1000} + \cos{\left((1.6714020808) \right)} + 3}{- \frac{1863 (1.6714020808)^{2}}{1000} - \sin{\left((1.6714020808) \right)}} = 1.6714002117$