Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{643 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{643 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{1929 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{643 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{1929 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4975813982$ $LaTeX: x_{2} = (2.4975813982) - \frac{- \frac{643 (2.4975813982)^{3}}{1000} + \sin{\left((2.4975813982) \right)} + 8}{- \frac{1929 (2.4975813982)^{2}}{1000} + \cos{\left((2.4975813982) \right)}} = 2.3871335672$ $LaTeX: x_{3} = (2.3871335672) - \frac{- \frac{643 (2.3871335672)^{3}}{1000} + \sin{\left((2.3871335672) \right)} + 8}{- \frac{1929 (2.3871335672)^{2}}{1000} + \cos{\left((2.3871335672) \right)}} = 2.3818657876$ $LaTeX: x_{4} = (2.3818657876) - \frac{- \frac{643 (2.3818657876)^{3}}{1000} + \sin{\left((2.3818657876) \right)} + 8}{- \frac{1929 (2.3818657876)^{2}}{1000} + \cos{\left((2.3818657876) \right)}} = 2.3818540292$ $LaTeX: x_{5} = (2.3818540292) - \frac{- \frac{643 (2.3818540292)^{3}}{1000} + \sin{\left((2.3818540292) \right)} + 8}{- \frac{1929 (2.3818540292)^{2}}{1000} + \cos{\left((2.3818540292) \right)}} = 2.3818540291$