Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{483 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{483 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 4}{- \frac{1449 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{483 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{1449 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 1.9558310040$ $LaTeX: x_{2} = (1.9558310040) - \frac{- \frac{483 (1.9558310040)^{3}}{500} + \cos{\left((1.9558310040) \right)} + 4}{- \frac{1449 (1.9558310040)^{2}}{500} - \sin{\left((1.9558310040) \right)}} = 1.6559076898$ $LaTeX: x_{3} = (1.6559076898) - \frac{- \frac{483 (1.6559076898)^{3}}{500} + \cos{\left((1.6559076898) \right)} + 4}{- \frac{1449 (1.6559076898)^{2}}{500} - \sin{\left((1.6559076898) \right)}} = 1.6032195112$ $LaTeX: x_{4} = (1.6032195112) - \frac{- \frac{483 (1.6032195112)^{3}}{500} + \cos{\left((1.6032195112) \right)} + 4}{- \frac{1449 (1.6032195112)^{2}}{500} - \sin{\left((1.6032195112) \right)}} = 1.6016704591$ $LaTeX: x_{5} = (1.6016704591) - \frac{- \frac{483 (1.6016704591)^{3}}{500} + \cos{\left((1.6016704591) \right)} + 4}{- \frac{1449 (1.6016704591)^{2}}{500} - \sin{\left((1.6016704591) \right)}} = 1.6016691422$