Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{347 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{347 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 8}{- \frac{1041 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{347 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{1041 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.4628505666$ $LaTeX: x_{2} = (2.4628505666) - \frac{- \frac{347 (2.4628505666)^{3}}{500} + \sin{\left((2.4628505666) \right)} + 8}{- \frac{1041 (2.4628505666)^{2}}{500} + \cos{\left((2.4628505666) \right)}} = 2.3330918639$ $LaTeX: x_{3} = (2.3330918639) - \frac{- \frac{347 (2.3330918639)^{3}}{500} + \sin{\left((2.3330918639) \right)} + 8}{- \frac{1041 (2.3330918639)^{2}}{500} + \cos{\left((2.3330918639) \right)}} = 2.3255748907$ $LaTeX: x_{4} = (2.3255748907) - \frac{- \frac{347 (2.3255748907)^{3}}{500} + \sin{\left((2.3255748907) \right)} + 8}{- \frac{1041 (2.3255748907)^{2}}{500} + \cos{\left((2.3255748907) \right)}} = 2.3255502231$ $LaTeX: x_{5} = (2.3255502231) - \frac{- \frac{347 (2.3255502231)^{3}}{500} + \sin{\left((2.3255502231) \right)} + 8}{- \frac{1041 (2.3255502231)^{2}}{500} + \cos{\left((2.3255502231) \right)}} = 2.3255502228$