Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{531 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{531 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 5}{- \frac{1593 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{531 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{1593 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.2867173023$ $LaTeX: x_{2} = (2.2867173023) - \frac{- \frac{531 (2.2867173023)^{3}}{1000} + \cos{\left((2.2867173023) \right)} + 5}{- \frac{1593 (2.2867173023)^{2}}{1000} - \sin{\left((2.2867173023) \right)}} = 2.0659321926$ $LaTeX: x_{3} = (2.0659321926) - \frac{- \frac{531 (2.0659321926)^{3}}{1000} + \cos{\left((2.0659321926) \right)} + 5}{- \frac{1593 (2.0659321926)^{2}}{1000} - \sin{\left((2.0659321926) \right)}} = 2.0454511292$ $LaTeX: x_{4} = (2.0454511292) - \frac{- \frac{531 (2.0454511292)^{3}}{1000} + \cos{\left((2.0454511292) \right)} + 5}{- \frac{1593 (2.0454511292)^{2}}{1000} - \sin{\left((2.0454511292) \right)}} = 2.0452820154$ $LaTeX: x_{5} = (2.0452820154) - \frac{- \frac{531 (2.0452820154)^{3}}{1000} + \cos{\left((2.0452820154) \right)} + 5}{- \frac{1593 (2.0452820154)^{2}}{1000} - \sin{\left((2.0452820154) \right)}} = 2.0452820039$