Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{357 x^{3}}{500} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{357 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 6}{- \frac{1071 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{357 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 6}{- \frac{1071 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 2.9528603883$ $LaTeX: x_{2} = (2.9528603883) - \frac{- \frac{357 (2.9528603883)^{3}}{500} + \cos{\left((2.9528603883) \right)} + 6}{- \frac{1071 (2.9528603883)^{2}}{500} - \sin{\left((2.9528603883) \right)}} = 2.2443515448$ $LaTeX: x_{3} = (2.2443515448) - \frac{- \frac{357 (2.2443515448)^{3}}{500} + \cos{\left((2.2443515448) \right)} + 6}{- \frac{1071 (2.2443515448)^{2}}{500} - \sin{\left((2.2443515448) \right)}} = 2.0113939296$ $LaTeX: x_{4} = (2.0113939296) - \frac{- \frac{357 (2.0113939296)^{3}}{500} + \cos{\left((2.0113939296) \right)} + 6}{- \frac{1071 (2.0113939296)^{2}}{500} - \sin{\left((2.0113939296) \right)}} = 1.9866655467$ $LaTeX: x_{5} = (1.9866655467) - \frac{- \frac{357 (1.9866655467)^{3}}{500} + \cos{\left((1.9866655467) \right)} + 6}{- \frac{1071 (1.9866655467)^{2}}{500} - \sin{\left((1.9866655467) \right)}} = 1.9863991701$