Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{803 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{803 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 2}{- \frac{2409 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{803 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{2409 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 2.0908511265$ $LaTeX: x_{2} = (2.0908511265) - \frac{- \frac{803 (2.0908511265)^{3}}{1000} + \sin{\left((2.0908511265) \right)} + 2}{- \frac{2409 (2.0908511265)^{2}}{1000} + \cos{\left((2.0908511265) \right)}} = 1.6853454552$ $LaTeX: x_{3} = (1.6853454552) - \frac{- \frac{803 (1.6853454552)^{3}}{1000} + \sin{\left((1.6853454552) \right)} + 2}{- \frac{2409 (1.6853454552)^{2}}{1000} + \cos{\left((1.6853454552) \right)}} = 1.5630845207$ $LaTeX: x_{4} = (1.5630845207) - \frac{- \frac{803 (1.5630845207)^{3}}{1000} + \sin{\left((1.5630845207) \right)} + 2}{- \frac{2409 (1.5630845207)^{2}}{1000} + \cos{\left((1.5630845207) \right)}} = 1.5517421806$ $LaTeX: x_{5} = (1.5517421806) - \frac{- \frac{803 (1.5517421806)^{3}}{1000} + \sin{\left((1.5517421806) \right)} + 2}{- \frac{2409 (1.5517421806)^{2}}{1000} + \cos{\left((1.5517421806) \right)}} = 1.5516474712$