Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{999 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{999 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 4}{- \frac{2997 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{999 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 4}{- \frac{2997 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 2.5640796969$ $LaTeX: x_{2} = (2.5640796969) - \frac{- \frac{999 (2.5640796969)^{3}}{1000} + \sin{\left((2.5640796969) \right)} + 4}{- \frac{2997 (2.5640796969)^{2}}{1000} + \cos{\left((2.5640796969) \right)}} = 1.9655505468$ $LaTeX: x_{3} = (1.9655505468) - \frac{- \frac{999 (1.9655505468)^{3}}{1000} + \sin{\left((1.9655505468) \right)} + 4}{- \frac{2997 (1.9655505468)^{2}}{1000} + \cos{\left((1.9655505468) \right)}} = 1.7429503384$ $LaTeX: x_{4} = (1.7429503384) - \frac{- \frac{999 (1.7429503384)^{3}}{1000} + \sin{\left((1.7429503384) \right)} + 4}{- \frac{2997 (1.7429503384)^{2}}{1000} + \cos{\left((1.7429503384) \right)}} = 1.7101388021$ $LaTeX: x_{5} = (1.7101388021) - \frac{- \frac{999 (1.7101388021)^{3}}{1000} + \sin{\left((1.7101388021) \right)} + 4}{- \frac{2997 (1.7101388021)^{2}}{1000} + \cos{\left((1.7101388021) \right)}} = 1.7094514866$