Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{269 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{269 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 8}{- \frac{807 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{269 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 8}{- \frac{807 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.4875618054$ $LaTeX: x_{2} = (2.4875618054) - \frac{- \frac{269 (2.4875618054)^{3}}{500} + \cos{\left((2.4875618054) \right)} + 8}{- \frac{807 (2.4875618054)^{2}}{500} - \sin{\left((2.4875618054) \right)}} = 2.3861023103$ $LaTeX: x_{3} = (2.3861023103) - \frac{- \frac{269 (2.3861023103)^{3}}{500} + \cos{\left((2.3861023103) \right)} + 8}{- \frac{807 (2.3861023103)^{2}}{500} - \sin{\left((2.3861023103) \right)}} = 2.3823764774$ $LaTeX: x_{4} = (2.3823764774) - \frac{- \frac{269 (2.3823764774)^{3}}{500} + \cos{\left((2.3823764774) \right)} + 8}{- \frac{807 (2.3823764774)^{2}}{500} - \sin{\left((2.3823764774) \right)}} = 2.3823715645$ $LaTeX: x_{5} = (2.3823715645) - \frac{- \frac{269 (2.3823715645)^{3}}{500} + \cos{\left((2.3823715645) \right)} + 8}{- \frac{807 (2.3823715645)^{2}}{500} - \sin{\left((2.3823715645) \right)}} = 2.3823715645$