Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{23 x^{3}}{100} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{23 x_{n}^{3}}{100} + \cos{\left(x_{n} \right)} + 6}{- \frac{69 x_{n}^{2}}{100} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{23 (3.0000000000)^{3}}{100} + \cos{\left((3.0000000000) \right)} + 6}{- \frac{69 (3.0000000000)^{2}}{100} - \sin{\left((3.0000000000) \right)}} = 2.8110581291$ $LaTeX: x_{2} = (2.8110581291) - \frac{- \frac{23 (2.8110581291)^{3}}{100} + \cos{\left((2.8110581291) \right)} + 6}{- \frac{69 (2.8110581291)^{2}}{100} - \sin{\left((2.8110581291) \right)}} = 2.8015573537$ $LaTeX: x_{3} = (2.8015573537) - \frac{- \frac{23 (2.8015573537)^{3}}{100} + \cos{\left((2.8015573537) \right)} + 6}{- \frac{69 (2.8015573537)^{2}}{100} - \sin{\left((2.8015573537) \right)}} = 2.8015343520$ $LaTeX: x_{4} = (2.8015343520) - \frac{- \frac{23 (2.8015343520)^{3}}{100} + \cos{\left((2.8015343520) \right)} + 6}{- \frac{69 (2.8015343520)^{2}}{100} - \sin{\left((2.8015343520) \right)}} = 2.8015343518$ $LaTeX: x_{5} = (2.8015343518) - \frac{- \frac{23 (2.8015343518)^{3}}{100} + \cos{\left((2.8015343518) \right)} + 6}{- \frac{69 (2.8015343518)^{2}}{100} - \sin{\left((2.8015343518) \right)}} = 2.8015343518$