Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{49 x^{3}}{500} - 1$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{49 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 1}{- \frac{147 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{49 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 1}{- \frac{147 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.5861157598$ $LaTeX: x_{2} = (2.5861157598) - \frac{- \frac{49 (2.5861157598)^{3}}{500} + \sin{\left((2.5861157598) \right)} + 1}{- \frac{147 (2.5861157598)^{2}}{500} + \cos{\left((2.5861157598) \right)}} = 2.5265783880$ $LaTeX: x_{3} = (2.5265783880) - \frac{- \frac{49 (2.5265783880)^{3}}{500} + \sin{\left((2.5265783880) \right)} + 1}{- \frac{147 (2.5265783880)^{2}}{500} + \cos{\left((2.5265783880) \right)}} = 2.5252275034$ $LaTeX: x_{4} = (2.5252275034) - \frac{- \frac{49 (2.5252275034)^{3}}{500} + \sin{\left((2.5252275034) \right)} + 1}{- \frac{147 (2.5252275034)^{2}}{500} + \cos{\left((2.5252275034) \right)}} = 2.5252268040$ $LaTeX: x_{5} = (2.5252268040) - \frac{- \frac{49 (2.5252268040)^{3}}{500} + \sin{\left((2.5252268040) \right)} + 1}{- \frac{147 (2.5252268040)^{2}}{500} + \cos{\left((2.5252268040) \right)}} = 2.5252268040$