Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{447 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{447 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{1341 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{447 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 5}{- \frac{1341 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 3.3100697164$ $LaTeX: x_{2} = (3.3100697164) - \frac{- \frac{447 (3.3100697164)^{3}}{500} + \sin{\left((3.3100697164) \right)} + 5}{- \frac{1341 (3.3100697164)^{2}}{500} + \cos{\left((3.3100697164) \right)}} = 2.4016355092$ $LaTeX: x_{3} = (2.4016355092) - \frac{- \frac{447 (2.4016355092)^{3}}{500} + \sin{\left((2.4016355092) \right)} + 5}{- \frac{1341 (2.4016355092)^{2}}{500} + \cos{\left((2.4016355092) \right)}} = 1.9876589110$ $LaTeX: x_{4} = (1.9876589110) - \frac{- \frac{447 (1.9876589110)^{3}}{500} + \sin{\left((1.9876589110) \right)} + 5}{- \frac{1341 (1.9876589110)^{2}}{500} + \cos{\left((1.9876589110) \right)}} = 1.8871166275$ $LaTeX: x_{5} = (1.8871166275) - \frac{- \frac{447 (1.8871166275)^{3}}{500} + \sin{\left((1.8871166275) \right)} + 5}{- \frac{1341 (1.8871166275)^{2}}{500} + \cos{\left((1.8871166275) \right)}} = 1.8812694408$