Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{557 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{557 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 3}{- \frac{1671 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{557 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 3}{- \frac{1671 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 3.9048179738$ $LaTeX: x_{2} = (3.9048179738) - \frac{- \frac{557 (3.9048179738)^{3}}{1000} + \sin{\left((3.9048179738) \right)} + 3}{- \frac{1671 (3.9048179738)^{2}}{1000} + \cos{\left((3.9048179738) \right)}} = 2.7272245796$ $LaTeX: x_{3} = (2.7272245796) - \frac{- \frac{557 (2.7272245796)^{3}}{1000} + \sin{\left((2.7272245796) \right)} + 3}{- \frac{1671 (2.7272245796)^{2}}{1000} + \cos{\left((2.7272245796) \right)}} = 2.1355056413$ $LaTeX: x_{4} = (2.1355056413) - \frac{- \frac{557 (2.1355056413)^{3}}{1000} + \sin{\left((2.1355056413) \right)} + 3}{- \frac{1671 (2.1355056413)^{2}}{1000} + \cos{\left((2.1355056413) \right)}} = 1.9418066153$ $LaTeX: x_{5} = (1.9418066153) - \frac{- \frac{557 (1.9418066153)^{3}}{1000} + \sin{\left((1.9418066153) \right)} + 3}{- \frac{1671 (1.9418066153)^{2}}{1000} + \cos{\left((1.9418066153) \right)}} = 1.9198531140$