Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{211 x^{3}}{250} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{211 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 2}{- \frac{633 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{211 (1.0000000000)^{3}}{250} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{633 (1.0000000000)^{2}}{250} + \cos{\left((1.0000000000) \right)}} = 2.0028986782$ $LaTeX: x_{2} = (2.0028986782) - \frac{- \frac{211 (2.0028986782)^{3}}{250} + \sin{\left((2.0028986782) \right)} + 2}{- \frac{633 (2.0028986782)^{2}}{250} + \cos{\left((2.0028986782) \right)}} = 1.6366681111$ $LaTeX: x_{3} = (1.6366681111) - \frac{- \frac{211 (1.6366681111)^{3}}{250} + \sin{\left((1.6366681111) \right)} + 2}{- \frac{633 (1.6366681111)^{2}}{250} + \cos{\left((1.6366681111) \right)}} = 1.5341074049$ $LaTeX: x_{4} = (1.5341074049) - \frac{- \frac{211 (1.5341074049)^{3}}{250} + \sin{\left((1.5341074049) \right)} + 2}{- \frac{633 (1.5341074049)^{2}}{250} + \cos{\left((1.5341074049) \right)}} = 1.5260135485$ $LaTeX: x_{5} = (1.5260135485) - \frac{- \frac{211 (1.5260135485)^{3}}{250} + \sin{\left((1.5260135485) \right)} + 2}{- \frac{633 (1.5260135485)^{2}}{250} + \cos{\left((1.5260135485) \right)}} = 1.5259645446$