Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{183 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{183 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 8}{- \frac{549 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{183 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{549 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.4399689605$ $LaTeX: x_{2} = (2.4399689605) - \frac{- \frac{183 (2.4399689605)^{3}}{250} + \sin{\left((2.4399689605) \right)} + 8}{- \frac{549 (2.4399689605)^{2}}{250} + \cos{\left((2.4399689605) \right)}} = 2.2963207881$ $LaTeX: x_{3} = (2.2963207881) - \frac{- \frac{183 (2.2963207881)^{3}}{250} + \sin{\left((2.2963207881) \right)} + 8}{- \frac{549 (2.2963207881)^{2}}{250} + \cos{\left((2.2963207881) \right)}} = 2.2868935396$ $LaTeX: x_{4} = (2.2868935396) - \frac{- \frac{183 (2.2868935396)^{3}}{250} + \sin{\left((2.2868935396) \right)} + 8}{- \frac{549 (2.2868935396)^{2}}{250} + \cos{\left((2.2868935396) \right)}} = 2.2868539320$ $LaTeX: x_{5} = (2.2868539320) - \frac{- \frac{183 (2.2868539320)^{3}}{250} + \sin{\left((2.2868539320) \right)} + 8}{- \frac{549 (2.2868539320)^{2}}{250} + \cos{\left((2.2868539320) \right)}} = 2.2868539313$