Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{113 x^{3}}{125} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{113 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 2}{- \frac{339 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{113 (1.0000000000)^{3}}{125} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{339 (1.0000000000)^{2}}{125} - \sin{\left((1.0000000000) \right)}} = 1.4604799963$ $LaTeX: x_{2} = (1.4604799963) - \frac{- \frac{113 (1.4604799963)^{3}}{125} + \cos{\left((1.4604799963) \right)} + 2}{- \frac{339 (1.4604799963)^{2}}{125} - \sin{\left((1.4604799963) \right)}} = 1.3563210794$ $LaTeX: x_{3} = (1.3563210794) - \frac{- \frac{113 (1.3563210794)^{3}}{125} + \cos{\left((1.3563210794) \right)} + 2}{- \frac{339 (1.3563210794)^{2}}{125} - \sin{\left((1.3563210794) \right)}} = 1.3491583657$ $LaTeX: x_{4} = (1.3491583657) - \frac{- \frac{113 (1.3491583657)^{3}}{125} + \cos{\left((1.3491583657) \right)} + 2}{- \frac{339 (1.3491583657)^{2}}{125} - \sin{\left((1.3491583657) \right)}} = 1.3491255675$ $LaTeX: x_{5} = (1.3491255675) - \frac{- \frac{113 (1.3491255675)^{3}}{125} + \cos{\left((1.3491255675) \right)} + 2}{- \frac{339 (1.3491255675)^{2}}{125} - \sin{\left((1.3491255675) \right)}} = 1.3491255669$