Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{141 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{141 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{423 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{141 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{423 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 3.9035077731$ $LaTeX: x_{2} = (3.9035077731) - \frac{- \frac{141 (3.9035077731)^{3}}{1000} + \sin{\left((3.9035077731) \right)} + 8}{- \frac{423 (3.9035077731)^{2}}{1000} + \cos{\left((3.9035077731) \right)}} = 3.7532933143$ $LaTeX: x_{3} = (3.7532933143) - \frac{- \frac{141 (3.7532933143)^{3}}{1000} + \sin{\left((3.7532933143) \right)} + 8}{- \frac{423 (3.7532933143)^{2}}{1000} + \cos{\left((3.7532933143) \right)}} = 3.7489532980$ $LaTeX: x_{4} = (3.7489532980) - \frac{- \frac{141 (3.7489532980)^{3}}{1000} + \sin{\left((3.7489532980) \right)} + 8}{- \frac{423 (3.7489532980)^{2}}{1000} + \cos{\left((3.7489532980) \right)}} = 3.7489496777$ $LaTeX: x_{5} = (3.7489496777) - \frac{- \frac{141 (3.7489496777)^{3}}{1000} + \sin{\left((3.7489496777) \right)} + 8}{- \frac{423 (3.7489496777)^{2}}{1000} + \cos{\left((3.7489496777) \right)}} = 3.7489496777$