Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{229 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{229 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 6}{- \frac{687 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{229 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 6}{- \frac{687 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.8145208353$ $LaTeX: x_{2} = (2.8145208353) - \frac{- \frac{229 (2.8145208353)^{3}}{1000} + \cos{\left((2.8145208353) \right)} + 6}{- \frac{687 (2.8145208353)^{2}}{1000} - \sin{\left((2.8145208353) \right)}} = 2.8053921643$ $LaTeX: x_{3} = (2.8053921643) - \frac{- \frac{229 (2.8053921643)^{3}}{1000} + \cos{\left((2.8053921643) \right)} + 6}{- \frac{687 (2.8053921643)^{2}}{1000} - \sin{\left((2.8053921643) \right)}} = 2.8053709782$ $LaTeX: x_{4} = (2.8053709782) - \frac{- \frac{229 (2.8053709782)^{3}}{1000} + \cos{\left((2.8053709782) \right)} + 6}{- \frac{687 (2.8053709782)^{2}}{1000} - \sin{\left((2.8053709782) \right)}} = 2.8053709781$ $LaTeX: x_{5} = (2.8053709781) - \frac{- \frac{229 (2.8053709781)^{3}}{1000} + \cos{\left((2.8053709781) \right)} + 6}{- \frac{687 (2.8053709781)^{2}}{1000} - \sin{\left((2.8053709781) \right)}} = 2.8053709781$