Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{487 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{487 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 7}{- \frac{1461 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{487 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{1461 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.5750842931$ $LaTeX: x_{2} = (2.5750842931) - \frac{- \frac{487 (2.5750842931)^{3}}{1000} + \sin{\left((2.5750842931) \right)} + 7}{- \frac{1461 (2.5750842931)^{2}}{1000} + \cos{\left((2.5750842931) \right)}} = 2.5011083233$ $LaTeX: x_{3} = (2.5011083233) - \frac{- \frac{487 (2.5011083233)^{3}}{1000} + \sin{\left((2.5011083233) \right)} + 7}{- \frac{1461 (2.5011083233)^{2}}{1000} + \cos{\left((2.5011083233) \right)}} = 2.4989037490$ $LaTeX: x_{4} = (2.4989037490) - \frac{- \frac{487 (2.4989037490)^{3}}{1000} + \sin{\left((2.4989037490) \right)} + 7}{- \frac{1461 (2.4989037490)^{2}}{1000} + \cos{\left((2.4989037490) \right)}} = 2.4989018134$ $LaTeX: x_{5} = (2.4989018134) - \frac{- \frac{487 (2.4989018134)^{3}}{1000} + \sin{\left((2.4989018134) \right)} + 7}{- \frac{1461 (2.4989018134)^{2}}{1000} + \cos{\left((2.4989018134) \right)}} = 2.4989018134$