Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{91 x^{3}}{125} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{91 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 8}{- \frac{273 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{91 (3.0000000000)^{3}}{125} + \cos{\left((3.0000000000) \right)} + 8}{- \frac{273 (3.0000000000)^{2}}{125} - \sin{\left((3.0000000000) \right)}} = 2.3612205972$ $LaTeX: x_{2} = (2.3612205972) - \frac{- \frac{91 (2.3612205972)^{3}}{125} + \cos{\left((2.3612205972) \right)} + 8}{- \frac{273 (2.3612205972)^{2}}{125} - \sin{\left((2.3612205972) \right)}} = 2.1830761098$ $LaTeX: x_{3} = (2.1830761098) - \frac{- \frac{91 (2.1830761098)^{3}}{125} + \cos{\left((2.1830761098) \right)} + 8}{- \frac{273 (2.1830761098)^{2}}{125} - \sin{\left((2.1830761098) \right)}} = 2.1698083039$ $LaTeX: x_{4} = (2.1698083039) - \frac{- \frac{91 (2.1698083039)^{3}}{125} + \cos{\left((2.1698083039) \right)} + 8}{- \frac{273 (2.1698083039)^{2}}{125} - \sin{\left((2.1698083039) \right)}} = 2.1697374257$ $LaTeX: x_{5} = (2.1697374257) - \frac{- \frac{91 (2.1697374257)^{3}}{125} + \cos{\left((2.1697374257) \right)} + 8}{- \frac{273 (2.1697374257)^{2}}{125} - \sin{\left((2.1697374257) \right)}} = 2.1697374237$