Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{211 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{211 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{633 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{211 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{633 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.4950836741$ $LaTeX: x_{2} = (2.4950836741) - \frac{- \frac{211 (2.4950836741)^{3}}{500} + \sin{\left((2.4950836741) \right)} + 5}{- \frac{633 (2.4950836741)^{2}}{500} + \cos{\left((2.4950836741) \right)}} = 2.3853410326$ $LaTeX: x_{3} = (2.3853410326) - \frac{- \frac{211 (2.3853410326)^{3}}{500} + \sin{\left((2.3853410326) \right)} + 5}{- \frac{633 (2.3853410326)^{2}}{500} + \cos{\left((2.3853410326) \right)}} = 2.3801354330$ $LaTeX: x_{4} = (2.3801354330) - \frac{- \frac{211 (2.3801354330)^{3}}{500} + \sin{\left((2.3801354330) \right)} + 5}{- \frac{633 (2.3801354330)^{2}}{500} + \cos{\left((2.3801354330) \right)}} = 2.3801238967$ $LaTeX: x_{5} = (2.3801238967) - \frac{- \frac{211 (2.3801238967)^{3}}{500} + \sin{\left((2.3801238967) \right)} + 5}{- \frac{633 (2.3801238967)^{2}}{500} + \cos{\left((2.3801238967) \right)}} = 2.3801238967$