Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{689 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{689 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{2067 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{689 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{2067 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.2926975412$ $LaTeX: x_{2} = (1.2926975412) - \frac{- \frac{689 (1.2926975412)^{3}}{1000} + \cos{\left((1.2926975412) \right)} + 1}{- \frac{2067 (1.2926975412)^{2}}{1000} - \sin{\left((1.2926975412) \right)}} = 1.2442703123$ $LaTeX: x_{3} = (1.2442703123) - \frac{- \frac{689 (1.2442703123)^{3}}{1000} + \cos{\left((1.2442703123) \right)} + 1}{- \frac{2067 (1.2442703123)^{2}}{1000} - \sin{\left((1.2442703123) \right)}} = 1.2426962376$ $LaTeX: x_{4} = (1.2426962376) - \frac{- \frac{689 (1.2426962376)^{3}}{1000} + \cos{\left((1.2426962376) \right)} + 1}{- \frac{2067 (1.2426962376)^{2}}{1000} - \sin{\left((1.2426962376) \right)}} = 1.2426946023$ $LaTeX: x_{5} = (1.2426946023) - \frac{- \frac{689 (1.2426946023)^{3}}{1000} + \cos{\left((1.2426946023) \right)} + 1}{- \frac{2067 (1.2426946023)^{2}}{1000} - \sin{\left((1.2426946023) \right)}} = 1.2426946023$