Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{353 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{353 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 8}{- \frac{1059 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{353 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 8}{- \frac{1059 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.3723940437$ $LaTeX: x_{2} = (2.3723940437) - \frac{- \frac{353 (2.3723940437)^{3}}{500} + \cos{\left((2.3723940437) \right)} + 8}{- \frac{1059 (2.3723940437)^{2}}{500} - \sin{\left((2.3723940437) \right)}} = 2.2023517665$ $LaTeX: x_{3} = (2.2023517665) - \frac{- \frac{353 (2.2023517665)^{3}}{500} + \cos{\left((2.2023517665) \right)} + 8}{- \frac{1059 (2.2023517665)^{2}}{500} - \sin{\left((2.2023517665) \right)}} = 2.1904365123$ $LaTeX: x_{4} = (2.1904365123) - \frac{- \frac{353 (2.1904365123)^{3}}{500} + \cos{\left((2.1904365123) \right)} + 8}{- \frac{1059 (2.1904365123)^{2}}{500} - \sin{\left((2.1904365123) \right)}} = 2.1903800844$ $LaTeX: x_{5} = (2.1903800844) - \frac{- \frac{353 (2.1903800844)^{3}}{500} + \cos{\left((2.1903800844) \right)} + 8}{- \frac{1059 (2.1903800844)^{2}}{500} - \sin{\left((2.1903800844) \right)}} = 2.1903800831$