Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{79 x^{3}}{100} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{79 x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 8}{- \frac{237 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{79 (3.0000000000)^{3}}{100} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{237 (3.0000000000)^{2}}{100} + \cos{\left((3.0000000000) \right)}} = 2.4091001601$ $LaTeX: x_{2} = (2.4091001601) - \frac{- \frac{79 (2.4091001601)^{3}}{100} + \sin{\left((2.4091001601) \right)} + 8}{- \frac{237 (2.4091001601)^{2}}{100} + \cos{\left((2.4091001601) \right)}} = 2.2451558640$ $LaTeX: x_{3} = (2.2451558640) - \frac{- \frac{79 (2.2451558640)^{3}}{100} + \sin{\left((2.2451558640) \right)} + 8}{- \frac{237 (2.2451558640)^{2}}{100} + \cos{\left((2.2451558640) \right)}} = 2.2324685436$ $LaTeX: x_{4} = (2.2324685436) - \frac{- \frac{79 (2.2324685436)^{3}}{100} + \sin{\left((2.2324685436) \right)} + 8}{- \frac{237 (2.2324685436)^{2}}{100} + \cos{\left((2.2324685436) \right)}} = 2.2323946698$ $LaTeX: x_{5} = (2.2323946698) - \frac{- \frac{79 (2.2323946698)^{3}}{100} + \sin{\left((2.2323946698) \right)} + 8}{- \frac{237 (2.2323946698)^{2}}{100} + \cos{\left((2.2323946698) \right)}} = 2.2323946673$