Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{417 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{417 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{1251 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{417 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{1251 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.5368305291$ $LaTeX: x_{2} = (1.5368305291) - \frac{- \frac{417 (1.5368305291)^{3}}{1000} + \cos{\left((1.5368305291) \right)} + 1}{- \frac{1251 (1.5368305291)^{2}}{1000} - \sin{\left((1.5368305291) \right)}} = 1.4155257228$ $LaTeX: x_{3} = (1.4155257228) - \frac{- \frac{417 (1.4155257228)^{3}}{1000} + \cos{\left((1.4155257228) \right)} + 1}{- \frac{1251 (1.4155257228)^{2}}{1000} - \sin{\left((1.4155257228) \right)}} = 1.4074868596$ $LaTeX: x_{4} = (1.4074868596) - \frac{- \frac{417 (1.4074868596)^{3}}{1000} + \cos{\left((1.4074868596) \right)} + 1}{- \frac{1251 (1.4074868596)^{2}}{1000} - \sin{\left((1.4074868596) \right)}} = 1.4074524285$ $LaTeX: x_{5} = (1.4074524285) - \frac{- \frac{417 (1.4074524285)^{3}}{1000} + \cos{\left((1.4074524285) \right)} + 1}{- \frac{1251 (1.4074524285)^{2}}{1000} - \sin{\left((1.4074524285) \right)}} = 1.4074524278$