Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{359 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{359 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 8}{- \frac{1077 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{359 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 8}{- \frac{1077 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.7271751316$ $LaTeX: x_{2} = (2.7271751316) - \frac{- \frac{359 (2.7271751316)^{3}}{1000} + \cos{\left((2.7271751316) \right)} + 8}{- \frac{1077 (2.7271751316)^{2}}{1000} - \sin{\left((2.7271751316) \right)}} = 2.7037509501$ $LaTeX: x_{3} = (2.7037509501) - \frac{- \frac{359 (2.7037509501)^{3}}{1000} + \cos{\left((2.7037509501) \right)} + 8}{- \frac{1077 (2.7037509501)^{2}}{1000} - \sin{\left((2.7037509501) \right)}} = 2.7035874315$ $LaTeX: x_{4} = (2.7035874315) - \frac{- \frac{359 (2.7035874315)^{3}}{1000} + \cos{\left((2.7035874315) \right)} + 8}{- \frac{1077 (2.7035874315)^{2}}{1000} - \sin{\left((2.7035874315) \right)}} = 2.7035874235$ $LaTeX: x_{5} = (2.7035874235) - \frac{- \frac{359 (2.7035874235)^{3}}{1000} + \cos{\left((2.7035874235) \right)} + 8}{- \frac{1077 (2.7035874235)^{2}}{1000} - \sin{\left((2.7035874235) \right)}} = 2.7035874235$