Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{763 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{763 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 2}{- \frac{2289 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{763 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{2289 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 2.1885822185$ $LaTeX: x_{2} = (2.1885822185) - \frac{- \frac{763 (2.1885822185)^{3}}{1000} + \sin{\left((2.1885822185) \right)} + 2}{- \frac{2289 (2.1885822185)^{2}}{1000} + \cos{\left((2.1885822185) \right)}} = 1.7395404455$ $LaTeX: x_{3} = (1.7395404455) - \frac{- \frac{763 (1.7395404455)^{3}}{1000} + \sin{\left((1.7395404455) \right)} + 2}{- \frac{2289 (1.7395404455)^{2}}{1000} + \cos{\left((1.7395404455) \right)}} = 1.5942831879$ $LaTeX: x_{4} = (1.5942831879) - \frac{- \frac{763 (1.5942831879)^{3}}{1000} + \sin{\left((1.5942831879) \right)} + 2}{- \frac{2289 (1.5942831879)^{2}}{1000} + \cos{\left((1.5942831879) \right)}} = 1.5785092674$ $LaTeX: x_{5} = (1.5785092674) - \frac{- \frac{763 (1.5785092674)^{3}}{1000} + \sin{\left((1.5785092674) \right)} + 2}{- \frac{2289 (1.5785092674)^{2}}{1000} + \cos{\left((1.5785092674) \right)}} = 1.5783290245$