Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{83 x^{3}}{125} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{83 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 8}{- \frac{249 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{83 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{249 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 2.4826681534$ $LaTeX: x_{2} = (2.4826681534) - \frac{- \frac{83 (2.4826681534)^{3}}{125} + \sin{\left((2.4826681534) \right)} + 8}{- \frac{249 (2.4826681534)^{2}}{125} + \cos{\left((2.4826681534) \right)}} = 2.3641825267$ $LaTeX: x_{3} = (2.3641825267) - \frac{- \frac{83 (2.3641825267)^{3}}{125} + \sin{\left((2.3641825267) \right)} + 8}{- \frac{249 (2.3641825267)^{2}}{125} + \cos{\left((2.3641825267) \right)}} = 2.3580343330$ $LaTeX: x_{4} = (2.3580343330) - \frac{- \frac{83 (2.3580343330)^{3}}{125} + \sin{\left((2.3580343330) \right)} + 8}{- \frac{249 (2.3580343330)^{2}}{125} + \cos{\left((2.3580343330) \right)}} = 2.3580181127$ $LaTeX: x_{5} = (2.3580181127) - \frac{- \frac{83 (2.3580181127)^{3}}{125} + \sin{\left((2.3580181127) \right)} + 8}{- \frac{249 (2.3580181127)^{2}}{125} + \cos{\left((2.3580181127) \right)}} = 2.3580181126$