Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{93 x^{3}}{125} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{93 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 1}{- \frac{279 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{93 (1.0000000000)^{3}}{125} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{279 (1.0000000000)^{2}}{125} - \sin{\left((1.0000000000) \right)}} = 1.2590889290$ $LaTeX: x_{2} = (1.2590889290) - \frac{- \frac{93 (1.2590889290)^{3}}{125} + \cos{\left((1.2590889290) \right)} + 1}{- \frac{279 (1.2590889290)^{2}}{125} - \sin{\left((1.2590889290) \right)}} = 1.2193648634$ $LaTeX: x_{3} = (1.2193648634) - \frac{- \frac{93 (1.2193648634)^{3}}{125} + \cos{\left((1.2193648634) \right)} + 1}{- \frac{279 (1.2193648634)^{2}}{125} - \sin{\left((1.2193648634) \right)}} = 1.2182750573$ $LaTeX: x_{4} = (1.2182750573) - \frac{- \frac{93 (1.2182750573)^{3}}{125} + \cos{\left((1.2182750573) \right)} + 1}{- \frac{279 (1.2182750573)^{2}}{125} - \sin{\left((1.2182750573) \right)}} = 1.2182742490$ $LaTeX: x_{5} = (1.2182742490) - \frac{- \frac{93 (1.2182742490)^{3}}{125} + \cos{\left((1.2182742490) \right)} + 1}{- \frac{279 (1.2182742490)^{2}}{125} - \sin{\left((1.2182742490) \right)}} = 1.2182742490$