Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{103 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{103 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 8}{- \frac{309 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{103 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{309 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.7537657389$ $LaTeX: x_{2} = (2.7537657389) - \frac{- \frac{103 (2.7537657389)^{3}}{250} + \sin{\left((2.7537657389) \right)} + 8}{- \frac{309 (2.7537657389)^{2}}{250} + \cos{\left((2.7537657389) \right)}} = 2.7318809768$ $LaTeX: x_{3} = (2.7318809768) - \frac{- \frac{103 (2.7318809768)^{3}}{250} + \sin{\left((2.7318809768) \right)} + 8}{- \frac{309 (2.7318809768)^{2}}{250} + \cos{\left((2.7318809768) \right)}} = 2.7317115761$ $LaTeX: x_{4} = (2.7317115761) - \frac{- \frac{103 (2.7317115761)^{3}}{250} + \sin{\left((2.7317115761) \right)} + 8}{- \frac{309 (2.7317115761)^{2}}{250} + \cos{\left((2.7317115761) \right)}} = 2.7317115660$ $LaTeX: x_{5} = (2.7317115660) - \frac{- \frac{103 (2.7317115660)^{3}}{250} + \sin{\left((2.7317115660) \right)} + 8}{- \frac{309 (2.7317115660)^{2}}{250} + \cos{\left((2.7317115660) \right)}} = 2.7317115660$