Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{569 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{569 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{1707 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{569 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{1707 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.3811313967$ $LaTeX: x_{2} = (1.3811313967) - \frac{- \frac{569 (1.3811313967)^{3}}{1000} + \cos{\left((1.3811313967) \right)} + 1}{- \frac{1707 (1.3811313967)^{2}}{1000} - \sin{\left((1.3811313967) \right)}} = 1.3078636642$ $LaTeX: x_{3} = (1.3078636642) - \frac{- \frac{569 (1.3078636642)^{3}}{1000} + \cos{\left((1.3078636642) \right)} + 1}{- \frac{1707 (1.3078636642)^{2}}{1000} - \sin{\left((1.3078636642) \right)}} = 1.3045172730$ $LaTeX: x_{4} = (1.3045172730) - \frac{- \frac{569 (1.3045172730)^{3}}{1000} + \cos{\left((1.3045172730) \right)} + 1}{- \frac{1707 (1.3045172730)^{2}}{1000} - \sin{\left((1.3045172730) \right)}} = 1.3045104402$ $LaTeX: x_{5} = (1.3045104402) - \frac{- \frac{569 (1.3045104402)^{3}}{1000} + \cos{\left((1.3045104402) \right)} + 1}{- \frac{1707 (1.3045104402)^{2}}{1000} - \sin{\left((1.3045104402) \right)}} = 1.3045104402$