Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{719 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{719 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{2157 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{719 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{2157 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.2739070380$ $LaTeX: x_{2} = (1.2739070380) - \frac{- \frac{719 (1.2739070380)^{3}}{1000} + \cos{\left((1.2739070380) \right)} + 1}{- \frac{2157 (1.2739070380)^{2}}{1000} - \sin{\left((1.2739070380) \right)}} = 1.2304052867$ $LaTeX: x_{3} = (1.2304052867) - \frac{- \frac{719 (1.2304052867)^{3}}{1000} + \cos{\left((1.2304052867) \right)} + 1}{- \frac{2157 (1.2304052867)^{2}}{1000} - \sin{\left((1.2304052867) \right)}} = 1.2291147591$ $LaTeX: x_{4} = (1.2291147591) - \frac{- \frac{719 (1.2291147591)^{3}}{1000} + \cos{\left((1.2291147591) \right)} + 1}{- \frac{2157 (1.2291147591)^{2}}{1000} - \sin{\left((1.2291147591) \right)}} = 1.2291136410$ $LaTeX: x_{5} = (1.2291136410) - \frac{- \frac{719 (1.2291136410)^{3}}{1000} + \cos{\left((1.2291136410) \right)} + 1}{- \frac{2157 (1.2291136410)^{2}}{1000} - \sin{\left((1.2291136410) \right)}} = 1.2291136410$