Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{919 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{919 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{2757 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{919 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{2757 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 1.4161464990$ $LaTeX: x_{2} = (1.4161464990) - \frac{- \frac{919 (1.4161464990)^{3}}{1000} + \sin{\left((1.4161464990) \right)} + 1}{- \frac{2757 (1.4161464990)^{2}}{1000} + \cos{\left((1.4161464990) \right)}} = 1.3004393186$ $LaTeX: x_{3} = (1.3004393186) - \frac{- \frac{919 (1.3004393186)^{3}}{1000} + \sin{\left((1.3004393186) \right)} + 1}{- \frac{2757 (1.3004393186)^{2}}{1000} + \cos{\left((1.3004393186) \right)}} = 1.2873768082$ $LaTeX: x_{4} = (1.2873768082) - \frac{- \frac{919 (1.2873768082)^{3}}{1000} + \sin{\left((1.2873768082) \right)} + 1}{- \frac{2757 (1.2873768082)^{2}}{1000} + \cos{\left((1.2873768082) \right)}} = 1.2872155301$ $LaTeX: x_{5} = (1.2872155301) - \frac{- \frac{919 (1.2872155301)^{3}}{1000} + \sin{\left((1.2872155301) \right)} + 1}{- \frac{2757 (1.2872155301)^{2}}{1000} + \cos{\left((1.2872155301) \right)}} = 1.2872155057$