Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{717 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{717 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 4}{- \frac{2151 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{717 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{2151 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.2776405604$ $LaTeX: x_{2} = (2.2776405604) - \frac{- \frac{717 (2.2776405604)^{3}}{1000} + \cos{\left((2.2776405604) \right)} + 4}{- \frac{2151 (2.2776405604)^{2}}{1000} - \sin{\left((2.2776405604) \right)}} = 1.8479736488$ $LaTeX: x_{3} = (1.8479736488) - \frac{- \frac{717 (1.8479736488)^{3}}{1000} + \cos{\left((1.8479736488) \right)} + 4}{- \frac{2151 (1.8479736488)^{2}}{1000} - \sin{\left((1.8479736488) \right)}} = 1.7518539225$ $LaTeX: x_{4} = (1.7518539225) - \frac{- \frac{717 (1.7518539225)^{3}}{1000} + \cos{\left((1.7518539225) \right)} + 4}{- \frac{2151 (1.7518539225)^{2}}{1000} - \sin{\left((1.7518539225) \right)}} = 1.7472438843$ $LaTeX: x_{5} = (1.7472438843) - \frac{- \frac{717 (1.7472438843)^{3}}{1000} + \cos{\left((1.7472438843) \right)} + 4}{- \frac{2151 (1.7472438843)^{2}}{1000} - \sin{\left((1.7472438843) \right)}} = 1.7472335393$