Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{481 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{481 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 8}{- \frac{1443 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{481 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{1443 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.3386409674$ $LaTeX: x_{2} = (2.3386409674) - \frac{- \frac{481 (2.3386409674)^{3}}{500} + \sin{\left((2.3386409674) \right)} + 8}{- \frac{1443 (2.3386409674)^{2}}{500} + \cos{\left((2.3386409674) \right)}} = 2.1210804244$ $LaTeX: x_{3} = (2.1210804244) - \frac{- \frac{481 (2.1210804244)^{3}}{500} + \sin{\left((2.1210804244) \right)} + 8}{- \frac{1443 (2.1210804244)^{2}}{500} + \cos{\left((2.1210804244) \right)}} = 2.0968185978$ $LaTeX: x_{4} = (2.0968185978) - \frac{- \frac{481 (2.0968185978)^{3}}{500} + \sin{\left((2.0968185978) \right)} + 8}{- \frac{1443 (2.0968185978)^{2}}{500} + \cos{\left((2.0968185978) \right)}} = 2.0965273603$ $LaTeX: x_{5} = (2.0965273603) - \frac{- \frac{481 (2.0965273603)^{3}}{500} + \sin{\left((2.0965273603) \right)} + 8}{- \frac{1443 (2.0965273603)^{2}}{500} + \cos{\left((2.0965273603) \right)}} = 2.0965273185$