Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{187 x^{3}}{250} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{187 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 2}{- \frac{561 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{187 (1.0000000000)^{3}}{250} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{561 (1.0000000000)^{2}}{250} + \cos{\left((1.0000000000) \right)}} = 2.2287807820$ $LaTeX: x_{2} = (2.2287807820) - \frac{- \frac{187 (2.2287807820)^{3}}{250} + \sin{\left((2.2287807820) \right)} + 2}{- \frac{561 (2.2287807820)^{2}}{250} + \cos{\left((2.2287807820) \right)}} = 1.7618703722$ $LaTeX: x_{3} = (1.7618703722) - \frac{- \frac{187 (1.7618703722)^{3}}{250} + \sin{\left((1.7618703722) \right)} + 2}{- \frac{561 (1.7618703722)^{2}}{250} + \cos{\left((1.7618703722) \right)}} = 1.6068693325$ $LaTeX: x_{4} = (1.6068693325) - \frac{- \frac{187 (1.6068693325)^{3}}{250} + \sin{\left((1.6068693325) \right)} + 2}{- \frac{561 (1.6068693325)^{2}}{250} + \cos{\left((1.6068693325) \right)}} = 1.5890155422$ $LaTeX: x_{5} = (1.5890155422) - \frac{- \frac{187 (1.5890155422)^{3}}{250} + \sin{\left((1.5890155422) \right)} + 2}{- \frac{561 (1.5890155422)^{2}}{250} + \cos{\left((1.5890155422) \right)}} = 1.5887860608$