Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{17 x^{3}}{50} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{50} + \cos{\left(x_{n} \right)} + 9}{- \frac{51 x_{n}^{2}}{50} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{17 (3.0000000000)^{3}}{50} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{51 (3.0000000000)^{2}}{50} - \sin{\left((3.0000000000) \right)}} = 2.8744794085$ $LaTeX: x_{2} = (2.8744794085) - \frac{- \frac{17 (2.8744794085)^{3}}{50} + \cos{\left((2.8744794085) \right)} + 9}{- \frac{51 (2.8744794085)^{2}}{50} - \sin{\left((2.8744794085) \right)}} = 2.8699007394$ $LaTeX: x_{3} = (2.8699007394) - \frac{- \frac{17 (2.8699007394)^{3}}{50} + \cos{\left((2.8699007394) \right)} + 9}{- \frac{51 (2.8699007394)^{2}}{50} - \sin{\left((2.8699007394) \right)}} = 2.8698948189$ $LaTeX: x_{4} = (2.8698948189) - \frac{- \frac{17 (2.8698948189)^{3}}{50} + \cos{\left((2.8698948189) \right)} + 9}{- \frac{51 (2.8698948189)^{2}}{50} - \sin{\left((2.8698948189) \right)}} = 2.8698948188$ $LaTeX: x_{5} = (2.8698948188) - \frac{- \frac{17 (2.8698948188)^{3}}{50} + \cos{\left((2.8698948188) \right)} + 9}{- \frac{51 (2.8698948188)^{2}}{50} - \sin{\left((2.8698948188) \right)}} = 2.8698948188$