Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{3 x^{3}}{250} - 7$ using $LaTeX: \displaystyle x_0=9$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{3 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 7}{- \frac{9 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 9$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (9.0000000000) - \frac{- \frac{3 (9.0000000000)^{3}}{250} + \sin{\left((9.0000000000) \right)} + 7}{- \frac{9 (9.0000000000)^{2}}{250} + \cos{\left((9.0000000000) \right)}} = 8.6509443308$ $LaTeX: x_{2} = (8.6509443308) - \frac{- \frac{3 (8.6509443308)^{3}}{250} + \sin{\left((8.6509443308) \right)} + 7}{- \frac{9 (8.6509443308)^{2}}{250} + \cos{\left((8.6509443308) \right)}} = 8.6303435217$ $LaTeX: x_{3} = (8.6303435217) - \frac{- \frac{3 (8.6303435217)^{3}}{250} + \sin{\left((8.6303435217) \right)} + 7}{- \frac{9 (8.6303435217)^{2}}{250} + \cos{\left((8.6303435217) \right)}} = 8.6302603173$ $LaTeX: x_{4} = (8.6302603173) - \frac{- \frac{3 (8.6302603173)^{3}}{250} + \sin{\left((8.6302603173) \right)} + 7}{- \frac{9 (8.6302603173)^{2}}{250} + \cos{\left((8.6302603173) \right)}} = 8.6302603159$ $LaTeX: x_{5} = (8.6302603159) - \frac{- \frac{3 (8.6302603159)^{3}}{250} + \sin{\left((8.6302603159) \right)} + 7}{- \frac{9 (8.6302603159)^{2}}{250} + \cos{\left((8.6302603159) \right)}} = 8.6302603159$