Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{609 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{609 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 2}{- \frac{1827 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{609 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{1827 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.7237486624$ $LaTeX: x_{2} = (1.7237486624) - \frac{- \frac{609 (1.7237486624)^{3}}{1000} + \cos{\left((1.7237486624) \right)} + 2}{- \frac{1827 (1.7237486624)^{2}}{1000} - \sin{\left((1.7237486624) \right)}} = 1.5255960314$ $LaTeX: x_{3} = (1.5255960314) - \frac{- \frac{609 (1.5255960314)^{3}}{1000} + \cos{\left((1.5255960314) \right)} + 2}{- \frac{1827 (1.5255960314)^{2}}{1000} - \sin{\left((1.5255960314) \right)}} = 1.5032746076$ $LaTeX: x_{4} = (1.5032746076) - \frac{- \frac{609 (1.5032746076)^{3}}{1000} + \cos{\left((1.5032746076) \right)} + 2}{- \frac{1827 (1.5032746076)^{2}}{1000} - \sin{\left((1.5032746076) \right)}} = 1.5030024738$ $LaTeX: x_{5} = (1.5030024738) - \frac{- \frac{609 (1.5030024738)^{3}}{1000} + \cos{\left((1.5030024738) \right)} + 2}{- \frac{1827 (1.5030024738)^{2}}{1000} - \sin{\left((1.5030024738) \right)}} = 1.5030024336$