Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{153 x^{3}}{250} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{153 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 9}{- \frac{459 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{153 (3.0000000000)^{3}}{250} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{459 (3.0000000000)^{2}}{250} - \sin{\left((3.0000000000) \right)}} = 2.4891130401$ $LaTeX: x_{2} = (2.4891130401) - \frac{- \frac{153 (2.4891130401)^{3}}{250} + \cos{\left((2.4891130401) \right)} + 9}{- \frac{459 (2.4891130401)^{2}}{250} - \sin{\left((2.4891130401) \right)}} = 2.3862377712$ $LaTeX: x_{3} = (2.3862377712) - \frac{- \frac{153 (2.3862377712)^{3}}{250} + \cos{\left((2.3862377712) \right)} + 9}{- \frac{459 (2.3862377712)^{2}}{250} - \sin{\left((2.3862377712) \right)}} = 2.3823231516$ $LaTeX: x_{4} = (2.3823231516) - \frac{- \frac{153 (2.3823231516)^{3}}{250} + \cos{\left((2.3823231516) \right)} + 9}{- \frac{459 (2.3823231516)^{2}}{250} - \sin{\left((2.3823231516) \right)}} = 2.3823176127$ $LaTeX: x_{5} = (2.3823176127) - \frac{- \frac{153 (2.3823176127)^{3}}{250} + \cos{\left((2.3823176127) \right)} + 9}{- \frac{459 (2.3823176127)^{2}}{250} - \sin{\left((2.3823176127) \right)}} = 2.3823176127$