Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{173 x^{3}}{250} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{173 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 3}{- \frac{519 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{173 (1.0000000000)^{3}}{250} + \cos{\left((1.0000000000) \right)} + 3}{- \frac{519 (1.0000000000)^{2}}{250} - \sin{\left((1.0000000000) \right)}} = 1.9762915623$ $LaTeX: x_{2} = (1.9762915623) - \frac{- \frac{173 (1.9762915623)^{3}}{250} + \cos{\left((1.9762915623) \right)} + 3}{- \frac{519 (1.9762915623)^{2}}{250} - \sin{\left((1.9762915623) \right)}} = 1.6732159862$ $LaTeX: x_{3} = (1.6732159862) - \frac{- \frac{173 (1.6732159862)^{3}}{250} + \cos{\left((1.6732159862) \right)} + 3}{- \frac{519 (1.6732159862)^{2}}{250} - \sin{\left((1.6732159862) \right)}} = 1.6226990190$ $LaTeX: x_{4} = (1.6226990190) - \frac{- \frac{173 (1.6226990190)^{3}}{250} + \cos{\left((1.6226990190) \right)} + 3}{- \frac{519 (1.6226990190)^{2}}{250} - \sin{\left((1.6226990190) \right)}} = 1.6213585519$ $LaTeX: x_{5} = (1.6213585519) - \frac{- \frac{173 (1.6213585519)^{3}}{250} + \cos{\left((1.6213585519) \right)} + 3}{- \frac{519 (1.6213585519)^{2}}{250} - \sin{\left((1.6213585519) \right)}} = 1.6213576218$