Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{999 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{999 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{2997 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{999 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{2997 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 1.3429282271$ $LaTeX: x_{2} = (1.3429282271) - \frac{- \frac{999 (1.3429282271)^{3}}{1000} + \sin{\left((1.3429282271) \right)} + 1}{- \frac{2997 (1.3429282271)^{2}}{1000} + \cos{\left((1.3429282271) \right)}} = 1.2569395724$ $LaTeX: x_{3} = (1.2569395724) - \frac{- \frac{999 (1.2569395724)^{3}}{1000} + \sin{\left((1.2569395724) \right)} + 1}{- \frac{2997 (1.2569395724)^{2}}{1000} + \cos{\left((1.2569395724) \right)}} = 1.2495519222$ $LaTeX: x_{4} = (1.2495519222) - \frac{- \frac{999 (1.2495519222)^{3}}{1000} + \sin{\left((1.2495519222) \right)} + 1}{- \frac{2997 (1.2495519222)^{2}}{1000} + \cos{\left((1.2495519222) \right)}} = 1.2494989563$ $LaTeX: x_{5} = (1.2494989563) - \frac{- \frac{999 (1.2494989563)^{3}}{1000} + \sin{\left((1.2494989563) \right)} + 1}{- \frac{2997 (1.2494989563)^{2}}{1000} + \cos{\left((1.2494989563) \right)}} = 1.2494989536$