Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{113 x^{3}}{125} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{113 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 5}{- \frac{339 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{113 (1.0000000000)^{3}}{125} + \sin{\left((1.0000000000) \right)} + 5}{- \frac{339 (1.0000000000)^{2}}{125} + \cos{\left((1.0000000000) \right)}} = 3.2735535421$ $LaTeX: x_{2} = (3.2735535421) - \frac{- \frac{113 (3.2735535421)^{3}}{125} + \sin{\left((3.2735535421) \right)} + 5}{- \frac{339 (3.2735535421)^{2}}{125} + \cos{\left((3.2735535421) \right)}} = 2.3803531858$ $LaTeX: x_{3} = (2.3803531858) - \frac{- \frac{113 (2.3803531858)^{3}}{125} + \sin{\left((2.3803531858) \right)} + 5}{- \frac{339 (2.3803531858)^{2}}{125} + \cos{\left((2.3803531858) \right)}} = 1.9762190826$ $LaTeX: x_{4} = (1.9762190826) - \frac{- \frac{113 (1.9762190826)^{3}}{125} + \sin{\left((1.9762190826) \right)} + 5}{- \frac{339 (1.9762190826)^{2}}{125} + \cos{\left((1.9762190826) \right)}} = 1.8799012720$ $LaTeX: x_{5} = (1.8799012720) - \frac{- \frac{113 (1.8799012720)^{3}}{125} + \sin{\left((1.8799012720) \right)} + 5}{- \frac{339 (1.8799012720)^{2}}{125} + \cos{\left((1.8799012720) \right)}} = 1.8745181492$