Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{161 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{161 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{483 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{161 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{483 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.6331182626$ $LaTeX: x_{2} = (2.6331182626) - \frac{- \frac{161 (2.6331182626)^{3}}{500} + \sin{\left((2.6331182626) \right)} + 5}{- \frac{483 (2.6331182626)^{2}}{500} + \cos{\left((2.6331182626) \right)}} = 2.5813870091$ $LaTeX: x_{3} = (2.5813870091) - \frac{- \frac{161 (2.5813870091)^{3}}{500} + \sin{\left((2.5813870091) \right)} + 5}{- \frac{483 (2.5813870091)^{2}}{500} + \cos{\left((2.5813870091) \right)}} = 2.5803664606$ $LaTeX: x_{4} = (2.5803664606) - \frac{- \frac{161 (2.5803664606)^{3}}{500} + \sin{\left((2.5803664606) \right)} + 5}{- \frac{483 (2.5803664606)^{2}}{500} + \cos{\left((2.5803664606) \right)}} = 2.5803660657$ $LaTeX: x_{5} = (2.5803660657) - \frac{- \frac{161 (2.5803660657)^{3}}{500} + \sin{\left((2.5803660657) \right)} + 5}{- \frac{483 (2.5803660657)^{2}}{500} + \cos{\left((2.5803660657) \right)}} = 2.5803660657$