Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{41 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{41 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 8}{- \frac{123 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{41 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{123 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 3.6853313308$ $LaTeX: x_{2} = (3.6853313308) - \frac{- \frac{41 (3.6853313308)^{3}}{250} + \sin{\left((3.6853313308) \right)} + 8}{- \frac{123 (3.6853313308)^{2}}{250} + \cos{\left((3.6853313308) \right)}} = 3.5890159257$ $LaTeX: x_{3} = (3.5890159257) - \frac{- \frac{41 (3.5890159257)^{3}}{250} + \sin{\left((3.5890159257) \right)} + 8}{- \frac{123 (3.5890159257)^{2}}{250} + \cos{\left((3.5890159257) \right)}} = 3.5870262417$ $LaTeX: x_{4} = (3.5870262417) - \frac{- \frac{41 (3.5870262417)^{3}}{250} + \sin{\left((3.5870262417) \right)} + 8}{- \frac{123 (3.5870262417)^{2}}{250} + \cos{\left((3.5870262417) \right)}} = 3.5870253936$ $LaTeX: x_{5} = (3.5870253936) - \frac{- \frac{41 (3.5870253936)^{3}}{250} + \sin{\left((3.5870253936) \right)} + 8}{- \frac{123 (3.5870253936)^{2}}{250} + \cos{\left((3.5870253936) \right)}} = 3.5870253936$