Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{331 x^{3}}{500} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{331 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 1}{- \frac{993 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{331 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{993 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 1.8158489770$ $LaTeX: x_{2} = (1.8158489770) - \frac{- \frac{331 (1.8158489770)^{3}}{500} + \sin{\left((1.8158489770) \right)} + 1}{- \frac{993 (1.8158489770)^{2}}{500} + \cos{\left((1.8158489770) \right)}} = 1.5222950303$ $LaTeX: x_{3} = (1.5222950303) - \frac{- \frac{331 (1.5222950303)^{3}}{500} + \sin{\left((1.5222950303) \right)} + 1}{- \frac{993 (1.5222950303)^{2}}{500} + \cos{\left((1.5222950303) \right)}} = 1.4483926753$ $LaTeX: x_{4} = (1.4483926753) - \frac{- \frac{331 (1.4483926753)^{3}}{500} + \sin{\left((1.4483926753) \right)} + 1}{- \frac{993 (1.4483926753)^{2}}{500} + \cos{\left((1.4483926753) \right)}} = 1.4437025987$ $LaTeX: x_{5} = (1.4437025987) - \frac{- \frac{331 (1.4437025987)^{3}}{500} + \sin{\left((1.4437025987) \right)} + 1}{- \frac{993 (1.4437025987)^{2}}{500} + \cos{\left((1.4437025987) \right)}} = 1.4436841271$