Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{89 x^{3}}{250} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{89 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 5}{- \frac{267 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{89 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{267 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.5782981366$ $LaTeX: x_{2} = (2.5782981366) - \frac{- \frac{89 (2.5782981366)^{3}}{250} + \sin{\left((2.5782981366) \right)} + 5}{- \frac{267 (2.5782981366)^{2}}{250} + \cos{\left((2.5782981366) \right)}} = 2.5068451974$ $LaTeX: x_{3} = (2.5068451974) - \frac{- \frac{89 (2.5068451974)^{3}}{250} + \sin{\left((2.5068451974) \right)} + 5}{- \frac{267 (2.5068451974)^{2}}{250} + \cos{\left((2.5068451974) \right)}} = 2.5048040783$ $LaTeX: x_{4} = (2.5048040783) - \frac{- \frac{89 (2.5048040783)^{3}}{250} + \sin{\left((2.5048040783) \right)} + 5}{- \frac{267 (2.5048040783)^{2}}{250} + \cos{\left((2.5048040783) \right)}} = 2.5048024277$ $LaTeX: x_{5} = (2.5048024277) - \frac{- \frac{89 (2.5048024277)^{3}}{250} + \sin{\left((2.5048024277) \right)} + 5}{- \frac{267 (2.5048024277)^{2}}{250} + \cos{\left((2.5048024277) \right)}} = 2.5048024277$