Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{153 x^{3}}{250} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{153 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 1}{- \frac{459 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{153 (1.0000000000)^{3}}{250} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{459 (1.0000000000)^{2}}{250} - \sin{\left((1.0000000000) \right)}} = 1.3467086333$ $LaTeX: x_{2} = (1.3467086333) - \frac{- \frac{153 (1.3467086333)^{3}}{250} + \cos{\left((1.3467086333) \right)} + 1}{- \frac{459 (1.3467086333)^{2}}{250} - \sin{\left((1.3467086333) \right)}} = 1.2833966586$ $LaTeX: x_{3} = (1.2833966586) - \frac{- \frac{153 (1.2833966586)^{3}}{250} + \cos{\left((1.2833966586) \right)} + 1}{- \frac{459 (1.2833966586)^{2}}{250} - \sin{\left((1.2833966586) \right)}} = 1.2808252394$ $LaTeX: x_{4} = (1.2808252394) - \frac{- \frac{153 (1.2808252394)^{3}}{250} + \cos{\left((1.2808252394) \right)} + 1}{- \frac{459 (1.2808252394)^{2}}{250} - \sin{\left((1.2808252394) \right)}} = 1.2808210810$ $LaTeX: x_{5} = (1.2808210810) - \frac{- \frac{153 (1.2808210810)^{3}}{250} + \cos{\left((1.2808210810) \right)} + 1}{- \frac{459 (1.2808210810)^{2}}{250} - \sin{\left((1.2808210810) \right)}} = 1.2808210810$