Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{189 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{189 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{567 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{189 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{567 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 3.4986252666$ $LaTeX: x_{2} = (3.4986252666) - \frac{- \frac{189 (3.4986252666)^{3}}{1000} + \sin{\left((3.4986252666) \right)} + 8}{- \frac{567 (3.4986252666)^{2}}{1000} + \cos{\left((3.4986252666) \right)}} = 3.4423458947$ $LaTeX: x_{3} = (3.4423458947) - \frac{- \frac{189 (3.4423458947)^{3}}{1000} + \sin{\left((3.4423458947) \right)} + 8}{- \frac{567 (3.4423458947)^{2}}{1000} + \cos{\left((3.4423458947) \right)}} = 3.4415999956$ $LaTeX: x_{4} = (3.4415999956) - \frac{- \frac{189 (3.4415999956)^{3}}{1000} + \sin{\left((3.4415999956) \right)} + 8}{- \frac{567 (3.4415999956)^{2}}{1000} + \cos{\left((3.4415999956) \right)}} = 3.4415998648$ $LaTeX: x_{5} = (3.4415998648) - \frac{- \frac{189 (3.4415998648)^{3}}{1000} + \sin{\left((3.4415998648) \right)} + 8}{- \frac{567 (3.4415998648)^{2}}{1000} + \cos{\left((3.4415998648) \right)}} = 3.4415998648$