Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{437 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{437 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{1311 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{437 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{1311 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.7139821614$ $LaTeX: x_{2} = (2.7139821614) - \frac{- \frac{437 (2.7139821614)^{3}}{1000} + \sin{\left((2.7139821614) \right)} + 8}{- \frac{1311 (2.7139821614)^{2}}{1000} + \cos{\left((2.7139821614) \right)}} = 2.6835936634$ $LaTeX: x_{3} = (2.6835936634) - \frac{- \frac{437 (2.6835936634)^{3}}{1000} + \sin{\left((2.6835936634) \right)} + 8}{- \frac{1311 (2.6835936634)^{2}}{1000} + \cos{\left((2.6835936634) \right)}} = 2.6832581008$ $LaTeX: x_{4} = (2.6832581008) - \frac{- \frac{437 (2.6832581008)^{3}}{1000} + \sin{\left((2.6832581008) \right)} + 8}{- \frac{1311 (2.6832581008)^{2}}{1000} + \cos{\left((2.6832581008) \right)}} = 2.6832580601$ $LaTeX: x_{5} = (2.6832580601) - \frac{- \frac{437 (2.6832580601)^{3}}{1000} + \sin{\left((2.6832580601) \right)} + 8}{- \frac{1311 (2.6832580601)^{2}}{1000} + \cos{\left((2.6832580601) \right)}} = 2.6832580601$