Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{69 x^{3}}{500} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{69 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 2}{- \frac{207 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{69 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 2}{- \frac{207 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.6639350056$ $LaTeX: x_{2} = (2.6639350056) - \frac{- \frac{69 (2.6639350056)^{3}}{500} + \sin{\left((2.6639350056) \right)} + 2}{- \frac{207 (2.6639350056)^{2}}{500} + \cos{\left((2.6639350056) \right)}} = 2.6249509295$ $LaTeX: x_{3} = (2.6249509295) - \frac{- \frac{69 (2.6249509295)^{3}}{500} + \sin{\left((2.6249509295) \right)} + 2}{- \frac{207 (2.6249509295)^{2}}{500} + \cos{\left((2.6249509295) \right)}} = 2.6244066233$ $LaTeX: x_{4} = (2.6244066233) - \frac{- \frac{69 (2.6244066233)^{3}}{500} + \sin{\left((2.6244066233) \right)} + 2}{- \frac{207 (2.6244066233)^{2}}{500} + \cos{\left((2.6244066233) \right)}} = 2.6244065171$ $LaTeX: x_{5} = (2.6244065171) - \frac{- \frac{69 (2.6244065171)^{3}}{500} + \sin{\left((2.6244065171) \right)} + 2}{- \frac{207 (2.6244065171)^{2}}{500} + \cos{\left((2.6244065171) \right)}} = 2.6244065171$