Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{819 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{819 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 2}{- \frac{2457 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{819 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{2457 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 2.0551851713$ $LaTeX: x_{2} = (2.0551851713) - \frac{- \frac{819 (2.0551851713)^{3}}{1000} + \sin{\left((2.0551851713) \right)} + 2}{- \frac{2457 (2.0551851713)^{2}}{1000} + \cos{\left((2.0551851713) \right)}} = 1.6655970306$ $LaTeX: x_{3} = (1.6655970306) - \frac{- \frac{819 (1.6655970306)^{3}}{1000} + \sin{\left((1.6655970306) \right)} + 2}{- \frac{2457 (1.6655970306)^{2}}{1000} + \cos{\left((1.6655970306) \right)}} = 1.5514496456$ $LaTeX: x_{4} = (1.5514496456) - \frac{- \frac{819 (1.5514496456)^{3}}{1000} + \sin{\left((1.5514496456) \right)} + 2}{- \frac{2457 (1.5514496456)^{2}}{1000} + \cos{\left((1.5514496456) \right)}} = 1.5415074083$ $LaTeX: x_{5} = (1.5415074083) - \frac{- \frac{819 (1.5415074083)^{3}}{1000} + \sin{\left((1.5415074083) \right)} + 2}{- \frac{2457 (1.5415074083)^{2}}{1000} + \cos{\left((1.5415074083) \right)}} = 1.5414341779$