Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{59 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{59 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 8}{- \frac{177 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{59 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{177 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 4.1865730152$ $LaTeX: x_{2} = (4.1865730152) - \frac{- \frac{59 (4.1865730152)^{3}}{500} + \sin{\left((4.1865730152) \right)} + 8}{- \frac{177 (4.1865730152)^{2}}{500} + \cos{\left((4.1865730152) \right)}} = 3.9593763526$ $LaTeX: x_{3} = (3.9593763526) - \frac{- \frac{59 (3.9593763526)^{3}}{500} + \sin{\left((3.9593763526) \right)} + 8}{- \frac{177 (3.9593763526)^{2}}{500} + \cos{\left((3.9593763526) \right)}} = 3.9507343970$ $LaTeX: x_{4} = (3.9507343970) - \frac{- \frac{59 (3.9507343970)^{3}}{500} + \sin{\left((3.9507343970) \right)} + 8}{- \frac{177 (3.9507343970)^{2}}{500} + \cos{\left((3.9507343970) \right)}} = 3.9507219395$ $LaTeX: x_{5} = (3.9507219395) - \frac{- \frac{59 (3.9507219395)^{3}}{500} + \sin{\left((3.9507219395) \right)} + 8}{- \frac{177 (3.9507219395)^{2}}{500} + \cos{\left((3.9507219395) \right)}} = 3.9507219394$