Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{719 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{719 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 7}{- \frac{2157 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{719 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{2157 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.3145693853$ $LaTeX: x_{2} = (2.3145693853) - \frac{- \frac{719 (2.3145693853)^{3}}{1000} + \cos{\left((2.3145693853) \right)} + 7}{- \frac{2157 (2.3145693853)^{2}}{1000} - \sin{\left((2.3145693853) \right)}} = 2.1036554223$ $LaTeX: x_{3} = (2.1036554223) - \frac{- \frac{719 (2.1036554223)^{3}}{1000} + \cos{\left((2.1036554223) \right)} + 7}{- \frac{2157 (2.1036554223)^{2}}{1000} - \sin{\left((2.1036554223) \right)}} = 2.0842942355$ $LaTeX: x_{4} = (2.0842942355) - \frac{- \frac{719 (2.0842942355)^{3}}{1000} + \cos{\left((2.0842942355) \right)} + 7}{- \frac{2157 (2.0842942355)^{2}}{1000} - \sin{\left((2.0842942355) \right)}} = 2.0841378591$ $LaTeX: x_{5} = (2.0841378591) - \frac{- \frac{719 (2.0841378591)^{3}}{1000} + \cos{\left((2.0841378591) \right)} + 7}{- \frac{2157 (2.0841378591)^{2}}{1000} - \sin{\left((2.0841378591) \right)}} = 2.0841378490$