Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{77 x^{3}}{250} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{77 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 9}{- \frac{231 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{77 (3.0000000000)^{3}}{250} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{231 (3.0000000000)^{2}}{250} - \sin{\left((3.0000000000) \right)}} = 2.9638183570$ $LaTeX: x_{2} = (2.9638183570) - \frac{- \frac{77 (2.9638183570)^{3}}{250} + \cos{\left((2.9638183570) \right)} + 9}{- \frac{231 (2.9638183570)^{2}}{250} - \sin{\left((2.9638183570) \right)}} = 2.9634605512$ $LaTeX: x_{3} = (2.9634605512) - \frac{- \frac{77 (2.9634605512)^{3}}{250} + \cos{\left((2.9634605512) \right)} + 9}{- \frac{231 (2.9634605512)^{2}}{250} - \sin{\left((2.9634605512) \right)}} = 2.9634605166$ $LaTeX: x_{4} = (2.9634605166) - \frac{- \frac{77 (2.9634605166)^{3}}{250} + \cos{\left((2.9634605166) \right)} + 9}{- \frac{231 (2.9634605166)^{2}}{250} - \sin{\left((2.9634605166) \right)}} = 2.9634605166$ $LaTeX: x_{5} = (2.9634605166) - \frac{- \frac{77 (2.9634605166)^{3}}{250} + \cos{\left((2.9634605166) \right)} + 9}{- \frac{231 (2.9634605166)^{2}}{250} - \sin{\left((2.9634605166) \right)}} = 2.9634605166$