Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{313 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{313 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{939 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{313 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{939 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.9671771806$ $LaTeX: x_{2} = (2.9671771806) - \frac{- \frac{313 (2.9671771806)^{3}}{1000} + \sin{\left((2.9671771806) \right)} + 8}{- \frac{939 (2.9671771806)^{2}}{1000} + \cos{\left((2.9671771806) \right)}} = 2.9668415057$ $LaTeX: x_{3} = (2.9668415057) - \frac{- \frac{313 (2.9668415057)^{3}}{1000} + \sin{\left((2.9668415057) \right)} + 8}{- \frac{939 (2.9668415057)^{2}}{1000} + \cos{\left((2.9668415057) \right)}} = 2.9668414707$ $LaTeX: x_{4} = (2.9668414707) - \frac{- \frac{313 (2.9668414707)^{3}}{1000} + \sin{\left((2.9668414707) \right)} + 8}{- \frac{939 (2.9668414707)^{2}}{1000} + \cos{\left((2.9668414707) \right)}} = 2.9668414707$ $LaTeX: x_{5} = (2.9668414707) - \frac{- \frac{313 (2.9668414707)^{3}}{1000} + \sin{\left((2.9668414707) \right)} + 8}{- \frac{939 (2.9668414707)^{2}}{1000} + \cos{\left((2.9668414707) \right)}} = 2.9668414707$