Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{229 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{229 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{687 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{229 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{687 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.4590533056$ $LaTeX: x_{2} = (2.4590533056) - \frac{- \frac{229 (2.4590533056)^{3}}{500} + \sin{\left((2.4590533056) \right)} + 5}{- \frac{687 (2.4590533056)^{2}}{500} + \cos{\left((2.4590533056) \right)}} = 2.3292073062$ $LaTeX: x_{3} = (2.3292073062) - \frac{- \frac{229 (2.3292073062)^{3}}{500} + \sin{\left((2.3292073062) \right)} + 5}{- \frac{687 (2.3292073062)^{2}}{500} + \cos{\left((2.3292073062) \right)}} = 2.3216470528$ $LaTeX: x_{4} = (2.3216470528) - \frac{- \frac{229 (2.3216470528)^{3}}{500} + \sin{\left((2.3216470528) \right)} + 5}{- \frac{687 (2.3216470528)^{2}}{500} + \cos{\left((2.3216470528) \right)}} = 2.3216218901$ $LaTeX: x_{5} = (2.3216218901) - \frac{- \frac{229 (2.3216218901)^{3}}{500} + \sin{\left((2.3216218901) \right)} + 5}{- \frac{687 (2.3216218901)^{2}}{500} + \cos{\left((2.3216218901) \right)}} = 2.3216218898$