Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{37 x^{3}}{250} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{37 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 5}{- \frac{111 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{37 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{111 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 3.2296674150$ $LaTeX: x_{2} = (3.2296674150) - \frac{- \frac{37 (3.2296674150)^{3}}{250} + \sin{\left((3.2296674150) \right)} + 5}{- \frac{111 (3.2296674150)^{2}}{250} + \cos{\left((3.2296674150) \right)}} = 3.2165593541$ $LaTeX: x_{3} = (3.2165593541) - \frac{- \frac{37 (3.2165593541)^{3}}{250} + \sin{\left((3.2165593541) \right)} + 5}{- \frac{111 (3.2165593541)^{2}}{250} + \cos{\left((3.2165593541) \right)}} = 3.2165166293$ $LaTeX: x_{4} = (3.2165166293) - \frac{- \frac{37 (3.2165166293)^{3}}{250} + \sin{\left((3.2165166293) \right)} + 5}{- \frac{111 (3.2165166293)^{2}}{250} + \cos{\left((3.2165166293) \right)}} = 3.2165166289$ $LaTeX: x_{5} = (3.2165166289) - \frac{- \frac{37 (3.2165166289)^{3}}{250} + \sin{\left((3.2165166289) \right)} + 5}{- \frac{111 (3.2165166289)^{2}}{250} + \cos{\left((3.2165166289) \right)}} = 3.2165166289$