Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{233 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{233 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 2}{- \frac{699 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{233 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{699 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.4977901750$ $LaTeX: x_{2} = (2.4977901750) - \frac{- \frac{233 (2.4977901750)^{3}}{1000} + \cos{\left((2.4977901750) \right)} + 2}{- \frac{699 (2.4977901750)^{2}}{1000} - \sin{\left((2.4977901750) \right)}} = 2.0078353974$ $LaTeX: x_{3} = (2.0078353974) - \frac{- \frac{233 (2.0078353974)^{3}}{1000} + \cos{\left((2.0078353974) \right)} + 2}{- \frac{699 (2.0078353974)^{2}}{1000} - \sin{\left((2.0078353974) \right)}} = 1.9247914216$ $LaTeX: x_{4} = (1.9247914216) - \frac{- \frac{233 (1.9247914216)^{3}}{1000} + \cos{\left((1.9247914216) \right)} + 2}{- \frac{699 (1.9247914216)^{2}}{1000} - \sin{\left((1.9247914216) \right)}} = 1.9224745386$ $LaTeX: x_{5} = (1.9224745386) - \frac{- \frac{233 (1.9224745386)^{3}}{1000} + \cos{\left((1.9224745386) \right)} + 2}{- \frac{699 (1.9224745386)^{2}}{1000} - \sin{\left((1.9224745386) \right)}} = 1.9224727525$