Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=11$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 6}{- \frac{3 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 11$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + \cos{\left((11.0000000000) \right)} + 6}{- \frac{3 (11.0000000000)^{2}}{1000} - \sin{\left((11.0000000000) \right)}} = 3.6632691524$ $LaTeX: x_{2} = (3.6632691524) - \frac{- \frac{(3.6632691524)^{3}}{1000} + \cos{\left((3.6632691524) \right)} + 6}{- \frac{3 (3.6632691524)^{2}}{1000} - \sin{\left((3.6632691524) \right)}} = -7.4350176849$ $LaTeX: x_{3} = (-7.4350176849) - \frac{- \frac{(-7.4350176849)^{3}}{1000} + \cos{\left((-7.4350176849) \right)} + 6}{- \frac{3 (-7.4350176849)^{2}}{1000} - \sin{\left((-7.4350176849) \right)}} = -16.5537410336$ $LaTeX: x_{4} = (-16.5537410336) - \frac{- \frac{(-16.5537410336)^{3}}{1000} + \cos{\left((-16.5537410336) \right)} + 6}{- \frac{3 (-16.5537410336)^{2}}{1000} - \sin{\left((-16.5537410336) \right)}} = -10.2674653278$ $LaTeX: x_{5} = (-10.2674653278) - \frac{- \frac{(-10.2674653278)^{3}}{1000} + \cos{\left((-10.2674653278) \right)} + 6}{- \frac{3 (-10.2674653278)^{2}}{1000} - \sin{\left((-10.2674653278) \right)}} = -4.2291049005$