Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{521 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{521 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 5}{- \frac{1563 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{521 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{1563 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.2921658537$ $LaTeX: x_{2} = (2.2921658537) - \frac{- \frac{521 (2.2921658537)^{3}}{1000} + \cos{\left((2.2921658537) \right)} + 5}{- \frac{1563 (2.2921658537)^{2}}{1000} - \sin{\left((2.2921658537) \right)}} = 2.0762918443$ $LaTeX: x_{3} = (2.0762918443) - \frac{- \frac{521 (2.0762918443)^{3}}{1000} + \cos{\left((2.0762918443) \right)} + 5}{- \frac{1563 (2.0762918443)^{2}}{1000} - \sin{\left((2.0762918443) \right)}} = 2.0568982343$ $LaTeX: x_{4} = (2.0568982343) - \frac{- \frac{521 (2.0568982343)^{3}}{1000} + \cos{\left((2.0568982343) \right)} + 5}{- \frac{1563 (2.0568982343)^{2}}{1000} - \sin{\left((2.0568982343) \right)}} = 2.0567479362$ $LaTeX: x_{5} = (2.0567479362) - \frac{- \frac{521 (2.0567479362)^{3}}{1000} + \cos{\left((2.0567479362) \right)} + 5}{- \frac{1563 (2.0567479362)^{2}}{1000} - \sin{\left((2.0567479362) \right)}} = 2.0567479272$