Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{823 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{823 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 2}{- \frac{2469 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{823 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{2469 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.5187486354$ $LaTeX: x_{2} = (1.5187486354) - \frac{- \frac{823 (1.5187486354)^{3}}{1000} + \cos{\left((1.5187486354) \right)} + 2}{- \frac{2469 (1.5187486354)^{2}}{1000} - \sin{\left((1.5187486354) \right)}} = 1.3945916674$ $LaTeX: x_{3} = (1.3945916674) - \frac{- \frac{823 (1.3945916674)^{3}}{1000} + \cos{\left((1.3945916674) \right)} + 2}{- \frac{2469 (1.3945916674)^{2}}{1000} - \sin{\left((1.3945916674) \right)}} = 1.3847502985$ $LaTeX: x_{4} = (1.3847502985) - \frac{- \frac{823 (1.3847502985)^{3}}{1000} + \cos{\left((1.3847502985) \right)} + 2}{- \frac{2469 (1.3847502985)^{2}}{1000} - \sin{\left((1.3847502985) \right)}} = 1.3846905923$ $LaTeX: x_{5} = (1.3846905923) - \frac{- \frac{823 (1.3846905923)^{3}}{1000} + \cos{\left((1.3846905923) \right)} + 2}{- \frac{2469 (1.3846905923)^{2}}{1000} - \sin{\left((1.3846905923) \right)}} = 1.3846905901$