Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{19 x^{3}}{125} - 1$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{19 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 1}{- \frac{57 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{19 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 1}{- \frac{57 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 2.4183579984$ $LaTeX: x_{2} = (2.4183579984) - \frac{- \frac{19 (2.4183579984)^{3}}{125} + \sin{\left((2.4183579984) \right)} + 1}{- \frac{57 (2.4183579984)^{2}}{125} + \cos{\left((2.4183579984) \right)}} = 2.2755177000$ $LaTeX: x_{3} = (2.2755177000) - \frac{- \frac{19 (2.2755177000)^{3}}{125} + \sin{\left((2.2755177000) \right)} + 1}{- \frac{57 (2.2755177000)^{2}}{125} + \cos{\left((2.2755177000) \right)}} = 2.2658263331$ $LaTeX: x_{4} = (2.2658263331) - \frac{- \frac{19 (2.2658263331)^{3}}{125} + \sin{\left((2.2658263331) \right)} + 1}{- \frac{57 (2.2658263331)^{2}}{125} + \cos{\left((2.2658263331) \right)}} = 2.2657816603$ $LaTeX: x_{5} = (2.2657816603) - \frac{- \frac{19 (2.2657816603)^{3}}{125} + \sin{\left((2.2657816603) \right)} + 1}{- \frac{57 (2.2657816603)^{2}}{125} + \cos{\left((2.2657816603) \right)}} = 2.2657816594$