Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{117 x^{3}}{125} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 2}{- \frac{351 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{117 (1.0000000000)^{3}}{125} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{351 (1.0000000000)^{2}}{125} + \cos{\left((1.0000000000) \right)}} = 1.8402667559$ $LaTeX: x_{2} = (1.8402667559) - \frac{- \frac{117 (1.8402667559)^{3}}{125} + \sin{\left((1.8402667559) \right)} + 2}{- \frac{351 (1.8402667559)^{2}}{125} + \cos{\left((1.8402667559) \right)}} = 1.5467401949$ $LaTeX: x_{3} = (1.5467401949) - \frac{- \frac{117 (1.5467401949)^{3}}{125} + \sin{\left((1.5467401949) \right)} + 2}{- \frac{351 (1.5467401949)^{2}}{125} + \cos{\left((1.5467401949) \right)}} = 1.4774387818$ $LaTeX: x_{4} = (1.4774387818) - \frac{- \frac{117 (1.4774387818)^{3}}{125} + \sin{\left((1.4774387818) \right)} + 2}{- \frac{351 (1.4774387818)^{2}}{125} + \cos{\left((1.4774387818) \right)}} = 1.4736373415$ $LaTeX: x_{5} = (1.4736373415) - \frac{- \frac{117 (1.4736373415)^{3}}{125} + \sin{\left((1.4736373415) \right)} + 2}{- \frac{351 (1.4736373415)^{2}}{125} + \cos{\left((1.4736373415) \right)}} = 1.4736261608$