Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{439 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{439 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 6}{- \frac{1317 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{439 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{1317 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.5552531862$ $LaTeX: x_{2} = (2.5552531862) - \frac{- \frac{439 (2.5552531862)^{3}}{1000} + \sin{\left((2.5552531862) \right)} + 6}{- \frac{1317 (2.5552531862)^{2}}{1000} + \cos{\left((2.5552531862) \right)}} = 2.4735121949$ $LaTeX: x_{3} = (2.4735121949) - \frac{- \frac{439 (2.4735121949)^{3}}{1000} + \sin{\left((2.4735121949) \right)} + 6}{- \frac{1317 (2.4735121949)^{2}}{1000} + \cos{\left((2.4735121949) \right)}} = 2.4707790155$ $LaTeX: x_{4} = (2.4707790155) - \frac{- \frac{439 (2.4707790155)^{3}}{1000} + \sin{\left((2.4707790155) \right)} + 6}{- \frac{1317 (2.4707790155)^{2}}{1000} + \cos{\left((2.4707790155) \right)}} = 2.4707759959$ $LaTeX: x_{5} = (2.4707759959) - \frac{- \frac{439 (2.4707759959)^{3}}{1000} + \sin{\left((2.4707759959) \right)} + 6}{- \frac{1317 (2.4707759959)^{2}}{1000} + \cos{\left((2.4707759959) \right)}} = 2.4707759959$