Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{93 x^{3}}{500} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{93 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 3}{- \frac{279 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{93 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 3}{- \frac{279 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.6871453195$ $LaTeX: x_{2} = (2.6871453195) - \frac{- \frac{93 (2.6871453195)^{3}}{500} + \sin{\left((2.6871453195) \right)} + 3}{- \frac{279 (2.6871453195)^{2}}{500} + \cos{\left((2.6871453195) \right)}} = 2.6526402081$ $LaTeX: x_{3} = (2.6526402081) - \frac{- \frac{93 (2.6526402081)^{3}}{500} + \sin{\left((2.6526402081) \right)} + 3}{- \frac{279 (2.6526402081)^{2}}{500} + \cos{\left((2.6526402081) \right)}} = 2.6522149759$ $LaTeX: x_{4} = (2.6522149759) - \frac{- \frac{93 (2.6522149759)^{3}}{500} + \sin{\left((2.6522149759) \right)} + 3}{- \frac{279 (2.6522149759)^{2}}{500} + \cos{\left((2.6522149759) \right)}} = 2.6522149114$ $LaTeX: x_{5} = (2.6522149114) - \frac{- \frac{93 (2.6522149114)^{3}}{500} + \sin{\left((2.6522149114) \right)} + 3}{- \frac{279 (2.6522149114)^{2}}{500} + \cos{\left((2.6522149114) \right)}} = 2.6522149114$