Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{131 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{131 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{393 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{131 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{393 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.7603073177$ $LaTeX: x_{2} = (2.7603073177) - \frac{- \frac{131 (2.7603073177)^{3}}{500} + \sin{\left((2.7603073177) \right)} + 5}{- \frac{393 (2.7603073177)^{2}}{500} + \cos{\left((2.7603073177) \right)}} = 2.7403324343$ $LaTeX: x_{3} = (2.7403324343) - \frac{- \frac{131 (2.7403324343)^{3}}{500} + \sin{\left((2.7403324343) \right)} + 5}{- \frac{393 (2.7403324343)^{2}}{500} + \cos{\left((2.7403324343) \right)}} = 2.7401948051$ $LaTeX: x_{4} = (2.7401948051) - \frac{- \frac{131 (2.7401948051)^{3}}{500} + \sin{\left((2.7401948051) \right)} + 5}{- \frac{393 (2.7401948051)^{2}}{500} + \cos{\left((2.7401948051) \right)}} = 2.7401947986$ $LaTeX: x_{5} = (2.7401947986) - \frac{- \frac{131 (2.7401947986)^{3}}{500} + \sin{\left((2.7401947986) \right)} + 5}{- \frac{393 (2.7401947986)^{2}}{500} + \cos{\left((2.7401947986) \right)}} = 2.7401947986$