Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{479 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{479 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 5}{- \frac{1437 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{479 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{1437 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.3175072211$ $LaTeX: x_{2} = (2.3175072211) - \frac{- \frac{479 (2.3175072211)^{3}}{1000} + \cos{\left((2.3175072211) \right)} + 5}{- \frac{1437 (2.3175072211)^{2}}{1000} - \sin{\left((2.3175072211) \right)}} = 2.1233098123$ $LaTeX: x_{3} = (2.1233098123) - \frac{- \frac{479 (2.1233098123)^{3}}{1000} + \cos{\left((2.1233098123) \right)} + 5}{- \frac{1437 (2.1233098123)^{2}}{1000} - \sin{\left((2.1233098123) \right)}} = 2.1082738348$ $LaTeX: x_{4} = (2.1082738348) - \frac{- \frac{479 (2.1082738348)^{3}}{1000} + \cos{\left((2.1082738348) \right)} + 5}{- \frac{1437 (2.1082738348)^{2}}{1000} - \sin{\left((2.1082738348) \right)}} = 2.1081869832$ $LaTeX: x_{5} = (2.1081869832) - \frac{- \frac{479 (2.1081869832)^{3}}{1000} + \cos{\left((2.1081869832) \right)} + 5}{- \frac{1437 (2.1081869832)^{2}}{1000} - \sin{\left((2.1081869832) \right)}} = 2.1081869803$