Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{121 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{121 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 2}{- \frac{363 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{121 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{363 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 2.4662487077$ $LaTeX: x_{2} = (2.4662487077) - \frac{- \frac{121 (2.4662487077)^{3}}{500} + \cos{\left((2.4662487077) \right)} + 2}{- \frac{363 (2.4662487077)^{2}}{500} - \sin{\left((2.4662487077) \right)}} = 1.9880369659$ $LaTeX: x_{3} = (1.9880369659) - \frac{- \frac{121 (1.9880369659)^{3}}{500} + \cos{\left((1.9880369659) \right)} + 2}{- \frac{363 (1.9880369659)^{2}}{500} - \sin{\left((1.9880369659) \right)}} = 1.9069744285$ $LaTeX: x_{4} = (1.9069744285) - \frac{- \frac{121 (1.9069744285)^{3}}{500} + \cos{\left((1.9069744285) \right)} + 2}{- \frac{363 (1.9069744285)^{2}}{500} - \sin{\left((1.9069744285) \right)}} = 1.9047128843$ $LaTeX: x_{5} = (1.9047128843) - \frac{- \frac{121 (1.9047128843)^{3}}{500} + \cos{\left((1.9047128843) \right)} + 2}{- \frac{363 (1.9047128843)^{2}}{500} - \sin{\left((1.9047128843) \right)}} = 1.9047111416$