Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{413 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{413 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 9}{- \frac{1239 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{413 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{1239 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.7218420904$ $LaTeX: x_{2} = (2.7218420904) - \frac{- \frac{413 (2.7218420904)^{3}}{1000} + \cos{\left((2.7218420904) \right)} + 9}{- \frac{1239 (2.7218420904)^{2}}{1000} - \sin{\left((2.7218420904) \right)}} = 2.6966866478$ $LaTeX: x_{3} = (2.6966866478) - \frac{- \frac{413 (2.6966866478)^{3}}{1000} + \cos{\left((2.6966866478) \right)} + 9}{- \frac{1239 (2.6966866478)^{2}}{1000} - \sin{\left((2.6966866478) \right)}} = 2.6964917849$ $LaTeX: x_{4} = (2.6964917849) - \frac{- \frac{413 (2.6964917849)^{3}}{1000} + \cos{\left((2.6964917849) \right)} + 9}{- \frac{1239 (2.6964917849)^{2}}{1000} - \sin{\left((2.6964917849) \right)}} = 2.6964917733$ $LaTeX: x_{5} = (2.6964917733) - \frac{- \frac{413 (2.6964917733)^{3}}{1000} + \cos{\left((2.6964917733) \right)} + 9}{- \frac{1239 (2.6964917733)^{2}}{1000} - \sin{\left((2.6964917733) \right)}} = 2.6964917733$