Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{309 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{309 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 2}{- \frac{927 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{309 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 2}{- \frac{927 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 2.6925286500$ $LaTeX: x_{2} = (2.6925286500) - \frac{- \frac{309 (2.6925286500)^{3}}{500} + \sin{\left((2.6925286500) \right)} + 2}{- \frac{927 (2.6925286500)^{2}}{500} + \cos{\left((2.6925286500) \right)}} = 2.0211164914$ $LaTeX: x_{3} = (2.0211164914) - \frac{- \frac{309 (2.0211164914)^{3}}{500} + \sin{\left((2.0211164914) \right)} + 2}{- \frac{927 (2.0211164914)^{2}}{500} + \cos{\left((2.0211164914) \right)}} = 1.7461709036$ $LaTeX: x_{4} = (1.7461709036) - \frac{- \frac{309 (1.7461709036)^{3}}{500} + \sin{\left((1.7461709036) \right)} + 2}{- \frac{927 (1.7461709036)^{2}}{500} + \cos{\left((1.7461709036) \right)}} = 1.6937063450$ $LaTeX: x_{5} = (1.6937063450) - \frac{- \frac{309 (1.6937063450)^{3}}{500} + \sin{\left((1.6937063450) \right)} + 2}{- \frac{927 (1.6937063450)^{2}}{500} + \cos{\left((1.6937063450) \right)}} = 1.6918352322$