Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{89 x^{3}}{100} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{89 x_{n}^{3}}{100} + \cos{\left(x_{n} \right)} + 5}{- \frac{267 x_{n}^{2}}{100} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{89 (1.0000000000)^{3}}{100} + \cos{\left((1.0000000000) \right)} + 5}{- \frac{267 (1.0000000000)^{2}}{100} - \sin{\left((1.0000000000) \right)}} = 2.3243174516$ $LaTeX: x_{2} = (2.3243174516) - \frac{- \frac{89 (2.3243174516)^{3}}{100} + \cos{\left((2.3243174516) \right)} + 5}{- \frac{267 (2.3243174516)^{2}}{100} - \sin{\left((2.3243174516) \right)}} = 1.8716297760$ $LaTeX: x_{3} = (1.8716297760) - \frac{- \frac{89 (1.8716297760)^{3}}{100} + \cos{\left((1.8716297760) \right)} + 5}{- \frac{267 (1.8716297760)^{2}}{100} - \sin{\left((1.8716297760) \right)}} = 1.7618677803$ $LaTeX: x_{4} = (1.7618677803) - \frac{- \frac{89 (1.7618677803)^{3}}{100} + \cos{\left((1.7618677803) \right)} + 5}{- \frac{267 (1.7618677803)^{2}}{100} - \sin{\left((1.7618677803) \right)}} = 1.7556697313$ $LaTeX: x_{5} = (1.7556697313) - \frac{- \frac{89 (1.7556697313)^{3}}{100} + \cos{\left((1.7556697313) \right)} + 5}{- \frac{267 (1.7556697313)^{2}}{100} - \sin{\left((1.7556697313) \right)}} = 1.7556505306$