Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{347 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{347 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 2}{- \frac{1041 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{347 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{1041 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 1.6315445973$ $LaTeX: x_{2} = (1.6315445973) - \frac{- \frac{347 (1.6315445973)^{3}}{500} + \cos{\left((1.6315445973) \right)} + 2}{- \frac{1041 (1.6315445973)^{2}}{500} - \sin{\left((1.6315445973) \right)}} = 1.4672096263$ $LaTeX: x_{3} = (1.4672096263) - \frac{- \frac{347 (1.4672096263)^{3}}{500} + \cos{\left((1.4672096263) \right)} + 2}{- \frac{1041 (1.4672096263)^{2}}{500} - \sin{\left((1.4672096263) \right)}} = 1.4510361161$ $LaTeX: x_{4} = (1.4510361161) - \frac{- \frac{347 (1.4510361161)^{3}}{500} + \cos{\left((1.4510361161) \right)} + 2}{- \frac{1041 (1.4510361161)^{2}}{500} - \sin{\left((1.4510361161) \right)}} = 1.4508853949$ $LaTeX: x_{5} = (1.4508853949) - \frac{- \frac{347 (1.4508853949)^{3}}{500} + \cos{\left((1.4508853949) \right)} + 2}{- \frac{1041 (1.4508853949)^{2}}{500} - \sin{\left((1.4508853949) \right)}} = 1.4508853819$