Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{133 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{133 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 5}{- \frac{399 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{133 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{399 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.5668523125$ $LaTeX: x_{2} = (2.5668523125) - \frac{- \frac{133 (2.5668523125)^{3}}{500} + \cos{\left((2.5668523125) \right)} + 5}{- \frac{399 (2.5668523125)^{2}}{500} - \sin{\left((2.5668523125) \right)}} = 2.5085898955$ $LaTeX: x_{3} = (2.5085898955) - \frac{- \frac{133 (2.5085898955)^{3}}{500} + \cos{\left((2.5085898955) \right)} + 5}{- \frac{399 (2.5085898955)^{2}}{500} - \sin{\left((2.5085898955) \right)}} = 2.5076111154$ $LaTeX: x_{4} = (2.5076111154) - \frac{- \frac{133 (2.5076111154)^{3}}{500} + \cos{\left((2.5076111154) \right)} + 5}{- \frac{399 (2.5076111154)^{2}}{500} - \sin{\left((2.5076111154) \right)}} = 2.5076108424$ $LaTeX: x_{5} = (2.5076108424) - \frac{- \frac{133 (2.5076108424)^{3}}{500} + \cos{\left((2.5076108424) \right)} + 5}{- \frac{399 (2.5076108424)^{2}}{500} - \sin{\left((2.5076108424) \right)}} = 2.5076108424$