Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{357 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{357 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 2}{- \frac{1071 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{357 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{1071 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 1.6121401264$ $LaTeX: x_{2} = (1.6121401264) - \frac{- \frac{357 (1.6121401264)^{3}}{500} + \cos{\left((1.6121401264) \right)} + 2}{- \frac{1071 (1.6121401264)^{2}}{500} - \sin{\left((1.6121401264) \right)}} = 1.4548263209$ $LaTeX: x_{3} = (1.4548263209) - \frac{- \frac{357 (1.4548263209)^{3}}{500} + \cos{\left((1.4548263209) \right)} + 2}{- \frac{1071 (1.4548263209)^{2}}{500} - \sin{\left((1.4548263209) \right)}} = 1.4398420577$ $LaTeX: x_{4} = (1.4398420577) - \frac{- \frac{357 (1.4398420577)^{3}}{500} + \cos{\left((1.4398420577) \right)} + 2}{- \frac{1071 (1.4398420577)^{2}}{500} - \sin{\left((1.4398420577) \right)}} = 1.4397112011$ $LaTeX: x_{5} = (1.4397112011) - \frac{- \frac{357 (1.4397112011)^{3}}{500} + \cos{\left((1.4397112011) \right)} + 2}{- \frac{1071 (1.4397112011)^{2}}{500} - \sin{\left((1.4397112011) \right)}} = 1.4397111912$