Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{116 x^{3}}{125} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{116 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 2}{- \frac{348 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{116 (1.0000000000)^{3}}{125} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{348 (1.0000000000)^{2}}{125} + \cos{\left((1.0000000000) \right)}} = 1.8528203197$ $LaTeX: x_{2} = (1.8528203197) - \frac{- \frac{116 (1.8528203197)^{3}}{125} + \sin{\left((1.8528203197) \right)} + 2}{- \frac{348 (1.8528203197)^{2}}{125} + \cos{\left((1.8528203197) \right)}} = 1.5536862179$ $LaTeX: x_{3} = (1.5536862179) - \frac{- \frac{116 (1.5536862179)^{3}}{125} + \sin{\left((1.5536862179) \right)} + 2}{- \frac{348 (1.5536862179)^{2}}{125} + \cos{\left((1.5536862179) \right)}} = 1.4819877416$ $LaTeX: x_{4} = (1.4819877416) - \frac{- \frac{116 (1.4819877416)^{3}}{125} + \sin{\left((1.4819877416) \right)} + 2}{- \frac{348 (1.4819877416)^{2}}{125} + \cos{\left((1.4819877416) \right)}} = 1.4779282543$ $LaTeX: x_{5} = (1.4779282543) - \frac{- \frac{116 (1.4779282543)^{3}}{125} + \sin{\left((1.4779282543) \right)} + 2}{- \frac{348 (1.4779282543)^{2}}{125} + \cos{\left((1.4779282543) \right)}} = 1.4779155401$