Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{489 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{489 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 8}{- \frac{1467 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{489 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 8}{- \frac{1467 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 3.0030089852$ $LaTeX: x_{2} = (3.0030089852) - \frac{- \frac{489 (3.0030089852)^{3}}{500} + \cos{\left((3.0030089852) \right)} + 8}{- \frac{1467 (3.0030089852)^{2}}{500} - \sin{\left((3.0030089852) \right)}} = 2.2707516957$ $LaTeX: x_{3} = (2.2707516957) - \frac{- \frac{489 (2.2707516957)^{3}}{500} + \cos{\left((2.2707516957) \right)} + 8}{- \frac{1467 (2.2707516957)^{2}}{500} - \sin{\left((2.2707516957) \right)}} = 2.0130802820$ $LaTeX: x_{4} = (2.0130802820) - \frac{- \frac{489 (2.0130802820)^{3}}{500} + \cos{\left((2.0130802820) \right)} + 8}{- \frac{1467 (2.0130802820)^{2}}{500} - \sin{\left((2.0130802820) \right)}} = 1.9813054003$ $LaTeX: x_{5} = (1.9813054003) - \frac{- \frac{489 (1.9813054003)^{3}}{500} + \cos{\left((1.9813054003) \right)} + 8}{- \frac{1467 (1.9813054003)^{2}}{500} - \sin{\left((1.9813054003) \right)}} = 1.9808453318$