Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{293 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{293 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 4}{- \frac{879 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{293 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{879 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 2.5211950158$ $LaTeX: x_{2} = (2.5211950158) - \frac{- \frac{293 (2.5211950158)^{3}}{500} + \cos{\left((2.5211950158) \right)} + 4}{- \frac{879 (2.5211950158)^{2}}{500} - \sin{\left((2.5211950158) \right)}} = 1.9933979331$ $LaTeX: x_{3} = (1.9933979331) - \frac{- \frac{293 (1.9933979331)^{3}}{500} + \cos{\left((1.9933979331) \right)} + 4}{- \frac{879 (1.9933979331)^{2}}{500} - \sin{\left((1.9933979331) \right)}} = 1.8602116730$ $LaTeX: x_{4} = (1.8602116730) - \frac{- \frac{293 (1.8602116730)^{3}}{500} + \cos{\left((1.8602116730) \right)} + 4}{- \frac{879 (1.8602116730)^{2}}{500} - \sin{\left((1.8602116730) \right)}} = 1.8520453922$ $LaTeX: x_{5} = (1.8520453922) - \frac{- \frac{293 (1.8520453922)^{3}}{500} + \cos{\left((1.8520453922) \right)} + 4}{- \frac{879 (1.8520453922)^{2}}{500} - \sin{\left((1.8520453922) \right)}} = 1.8520155902$