Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{23 x^{3}}{500} - 1$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{23 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 1}{- \frac{69 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{23 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 1}{- \frac{69 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.1092656534$ $LaTeX: x_{2} = (2.1092656534) - \frac{- \frac{23 (2.1092656534)^{3}}{500} + \cos{\left((2.1092656534) \right)} + 1}{- \frac{69 (2.1092656534)^{2}}{500} - \sin{\left((2.1092656534) \right)}} = 2.1469629708$ $LaTeX: x_{3} = (2.1469629708) - \frac{- \frac{23 (2.1469629708)^{3}}{500} + \cos{\left((2.1469629708) \right)} + 1}{- \frac{69 (2.1469629708)^{2}}{500} - \sin{\left((2.1469629708) \right)}} = 2.1469330600$ $LaTeX: x_{4} = (2.1469330600) - \frac{- \frac{23 (2.1469330600)^{3}}{500} + \cos{\left((2.1469330600) \right)} + 1}{- \frac{69 (2.1469330600)^{2}}{500} - \sin{\left((2.1469330600) \right)}} = 2.1469330600$ $LaTeX: x_{5} = (2.1469330600) - \frac{- \frac{23 (2.1469330600)^{3}}{500} + \cos{\left((2.1469330600) \right)} + 1}{- \frac{69 (2.1469330600)^{2}}{500} - \sin{\left((2.1469330600) \right)}} = 2.1469330600$