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

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{293 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 1}{- \frac{879 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{293 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{879 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 1.3671140441$ $LaTeX: x_{2} = (1.3671140441) - \frac{- \frac{293 (1.3671140441)^{3}}{500} + \cos{\left((1.3671140441) \right)} + 1}{- \frac{879 (1.3671140441)^{2}}{500} - \sin{\left((1.3671140441) \right)}} = 1.2979390444$ $LaTeX: x_{3} = (1.2979390444) - \frac{- \frac{293 (1.2979390444)^{3}}{500} + \cos{\left((1.2979390444) \right)} + 1}{- \frac{879 (1.2979390444)^{2}}{500} - \sin{\left((1.2979390444) \right)}} = 1.2949210513$ $LaTeX: x_{4} = (1.2949210513) - \frac{- \frac{293 (1.2949210513)^{3}}{500} + \cos{\left((1.2949210513) \right)} + 1}{- \frac{879 (1.2949210513)^{2}}{500} - \sin{\left((1.2949210513) \right)}} = 1.2949154251$ $LaTeX: x_{5} = (1.2949154251) - \frac{- \frac{293 (1.2949154251)^{3}}{500} + \cos{\left((1.2949154251) \right)} + 1}{- \frac{879 (1.2949154251)^{2}}{500} - \sin{\left((1.2949154251) \right)}} = 1.2949154251$