Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{483 x^{3}}{500} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{483 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 1}{- \frac{1449 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{483 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{1449 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 1.1535784896$ $LaTeX: x_{2} = (1.1535784896) - \frac{- \frac{483 (1.1535784896)^{3}}{500} + \cos{\left((1.1535784896) \right)} + 1}{- \frac{1449 (1.1535784896)^{2}}{500} - \sin{\left((1.1535784896) \right)}} = 1.1372906987$ $LaTeX: x_{3} = (1.1372906987) - \frac{- \frac{483 (1.1372906987)^{3}}{500} + \cos{\left((1.1372906987) \right)} + 1}{- \frac{1449 (1.1372906987)^{2}}{500} - \sin{\left((1.1372906987) \right)}} = 1.1370894203$ $LaTeX: x_{4} = (1.1370894203) - \frac{- \frac{483 (1.1370894203)^{3}}{500} + \cos{\left((1.1370894203) \right)} + 1}{- \frac{1449 (1.1370894203)^{2}}{500} - \sin{\left((1.1370894203) \right)}} = 1.1370893898$ $LaTeX: x_{5} = (1.1370893898) - \frac{- \frac{483 (1.1370893898)^{3}}{500} + \cos{\left((1.1370893898) \right)} + 1}{- \frac{1449 (1.1370893898)^{2}}{500} - \sin{\left((1.1370893898) \right)}} = 1.1370893898$