Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{81 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{81 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 7}{- \frac{243 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{81 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{243 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 2.4398526347$ $LaTeX: x_{2} = (2.4398526347) - \frac{- \frac{81 (2.4398526347)^{3}}{125} + \sin{\left((2.4398526347) \right)} + 7}{- \frac{243 (2.4398526347)^{2}}{125} + \cos{\left((2.4398526347) \right)}} = 2.2966874470$ $LaTeX: x_{3} = (2.2966874470) - \frac{- \frac{81 (2.2966874470)^{3}}{125} + \sin{\left((2.2966874470) \right)} + 7}{- \frac{243 (2.2966874470)^{2}}{125} + \cos{\left((2.2966874470) \right)}} = 2.2873183584$ $LaTeX: x_{4} = (2.2873183584) - \frac{- \frac{81 (2.2873183584)^{3}}{125} + \sin{\left((2.2873183584) \right)} + 7}{- \frac{243 (2.2873183584)^{2}}{125} + \cos{\left((2.2873183584) \right)}} = 2.2872791709$ $LaTeX: x_{5} = (2.2872791709) - \frac{- \frac{81 (2.2872791709)^{3}}{125} + \sin{\left((2.2872791709) \right)} + 7}{- \frac{243 (2.2872791709)^{2}}{125} + \cos{\left((2.2872791709) \right)}} = 2.2872791703$