Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{523 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{523 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 5}{- \frac{1569 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{523 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{1569 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.2910596397$ $LaTeX: x_{2} = (2.2910596397) - \frac{- \frac{523 (2.2910596397)^{3}}{1000} + \cos{\left((2.2910596397) \right)} + 5}{- \frac{1569 (2.2910596397)^{2}}{1000} - \sin{\left((2.2910596397) \right)}} = 2.0741959184$ $LaTeX: x_{3} = (2.0741959184) - \frac{- \frac{523 (2.0741959184)^{3}}{1000} + \cos{\left((2.0741959184) \right)} + 5}{- \frac{1569 (2.0741959184)^{2}}{1000} - \sin{\left((2.0741959184) \right)}} = 2.0545862270$ $LaTeX: x_{4} = (2.0545862270) - \frac{- \frac{523 (2.0545862270)^{3}}{1000} + \cos{\left((2.0545862270) \right)} + 5}{- \frac{1569 (2.0545862270)^{2}}{1000} - \sin{\left((2.0545862270) \right)}} = 2.0544322867$ $LaTeX: x_{5} = (2.0544322867) - \frac{- \frac{523 (2.0544322867)^{3}}{1000} + \cos{\left((2.0544322867) \right)} + 5}{- \frac{1569 (2.0544322867)^{2}}{1000} - \sin{\left((2.0544322867) \right)}} = 2.0544322773$