Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{329 x^{3}}{500} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{329 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 9}{- \frac{987 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{329 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{987 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.4551891933$ $LaTeX: x_{2} = (2.4551891933) - \frac{- \frac{329 (2.4551891933)^{3}}{500} + \cos{\left((2.4551891933) \right)} + 9}{- \frac{987 (2.4551891933)^{2}}{500} - \sin{\left((2.4551891933) \right)}} = 2.3345648792$ $LaTeX: x_{3} = (2.3345648792) - \frac{- \frac{329 (2.3345648792)^{3}}{500} + \cos{\left((2.3345648792) \right)} + 9}{- \frac{987 (2.3345648792)^{2}}{500} - \sin{\left((2.3345648792) \right)}} = 2.3289966835$ $LaTeX: x_{4} = (2.3289966835) - \frac{- \frac{329 (2.3289966835)^{3}}{500} + \cos{\left((2.3289966835) \right)} + 9}{- \frac{987 (2.3289966835)^{2}}{500} - \sin{\left((2.3289966835) \right)}} = 2.3289851324$ $LaTeX: x_{5} = (2.3289851324) - \frac{- \frac{329 (2.3289851324)^{3}}{500} + \cos{\left((2.3289851324) \right)} + 9}{- \frac{987 (2.3289851324)^{2}}{500} - \sin{\left((2.3289851324) \right)}} = 2.3289851324$