Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{89 x^{3}}{200} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{89 x_{n}^{3}}{200} + \sin{\left(x_{n} \right)} + 8}{- \frac{267 x_{n}^{2}}{200} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{89 (3.0000000000)^{3}}{200} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{267 (3.0000000000)^{2}}{200} + \cos{\left((3.0000000000) \right)}} = 2.7021236273$ $LaTeX: x_{2} = (2.7021236273) - \frac{- \frac{89 (2.7021236273)^{3}}{200} + \sin{\left((2.7021236273) \right)} + 8}{- \frac{267 (2.7021236273)^{2}}{200} + \cos{\left((2.7021236273) \right)}} = 2.6688768204$ $LaTeX: x_{3} = (2.6688768204) - \frac{- \frac{89 (2.6688768204)^{3}}{200} + \sin{\left((2.6688768204) \right)} + 8}{- \frac{267 (2.6688768204)^{2}}{200} + \cos{\left((2.6688768204) \right)}} = 2.6684718286$ $LaTeX: x_{4} = (2.6684718286) - \frac{- \frac{89 (2.6684718286)^{3}}{200} + \sin{\left((2.6684718286) \right)} + 8}{- \frac{267 (2.6684718286)^{2}}{200} + \cos{\left((2.6684718286) \right)}} = 2.6684717688$ $LaTeX: x_{5} = (2.6684717688) - \frac{- \frac{89 (2.6684717688)^{3}}{200} + \sin{\left((2.6684717688) \right)} + 8}{- \frac{267 (2.6684717688)^{2}}{200} + \cos{\left((2.6684717688) \right)}} = 2.6684717688$