Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{199 x^{3}}{500} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{199 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 3}{- \frac{597 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{199 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 3}{- \frac{597 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.3520036764$ $LaTeX: x_{2} = (2.3520036764) - \frac{- \frac{199 (2.3520036764)^{3}}{500} + \sin{\left((2.3520036764) \right)} + 3}{- \frac{597 (2.3520036764)^{2}}{500} + \cos{\left((2.3520036764) \right)}} = 2.1511139377$ $LaTeX: x_{3} = (2.1511139377) - \frac{- \frac{199 (2.1511139377)^{3}}{500} + \sin{\left((2.1511139377) \right)} + 3}{- \frac{597 (2.1511139377)^{2}}{500} + \cos{\left((2.1511139377) \right)}} = 2.1304766376$ $LaTeX: x_{4} = (2.1304766376) - \frac{- \frac{199 (2.1304766376)^{3}}{500} + \sin{\left((2.1304766376) \right)} + 3}{- \frac{597 (2.1304766376)^{2}}{500} + \cos{\left((2.1304766376) \right)}} = 2.1302633283$ $LaTeX: x_{5} = (2.1302633283) - \frac{- \frac{199 (2.1302633283)^{3}}{500} + \sin{\left((2.1302633283) \right)} + 3}{- \frac{597 (2.1302633283)^{2}}{500} + \cos{\left((2.1302633283) \right)}} = 2.1302633056$