Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{411 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{411 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{1233 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{411 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 5}{- \frac{1233 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 3.6065726724$ $LaTeX: x_{2} = (3.6065726724) - \frac{- \frac{411 (3.6065726724)^{3}}{500} + \sin{\left((3.6065726724) \right)} + 5}{- \frac{1233 (3.6065726724)^{2}}{500} + \cos{\left((3.6065726724) \right)}} = 2.5750263260$ $LaTeX: x_{3} = (2.5750263260) - \frac{- \frac{411 (2.5750263260)^{3}}{500} + \sin{\left((2.5750263260) \right)} + 5}{- \frac{1233 (2.5750263260)^{2}}{500} + \cos{\left((2.5750263260) \right)}} = 2.0807952280$ $LaTeX: x_{4} = (2.0807952280) - \frac{- \frac{411 (2.0807952280)^{3}}{500} + \sin{\left((2.0807952280) \right)} + 5}{- \frac{1233 (2.0807952280)^{2}}{500} + \cos{\left((2.0807952280) \right)}} = 1.9435076884$ $LaTeX: x_{5} = (1.9435076884) - \frac{- \frac{411 (1.9435076884)^{3}}{500} + \sin{\left((1.9435076884) \right)} + 5}{- \frac{1233 (1.9435076884)^{2}}{500} + \cos{\left((1.9435076884) \right)}} = 1.9328650070$