Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{713 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{713 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{2139 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{713 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{2139 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4511198009$ $LaTeX: x_{2} = (2.4511198009) - \frac{- \frac{713 (2.4511198009)^{3}}{1000} + \sin{\left((2.4511198009) \right)} + 8}{- \frac{2139 (2.4511198009)^{2}}{1000} + \cos{\left((2.4511198009) \right)}} = 2.3143597763$ $LaTeX: x_{3} = (2.3143597763) - \frac{- \frac{713 (2.3143597763)^{3}}{1000} + \sin{\left((2.3143597763) \right)} + 8}{- \frac{2139 (2.3143597763)^{2}}{1000} + \cos{\left((2.3143597763) \right)}} = 2.3059114360$ $LaTeX: x_{4} = (2.3059114360) - \frac{- \frac{713 (2.3059114360)^{3}}{1000} + \sin{\left((2.3059114360) \right)} + 8}{- \frac{2139 (2.3059114360)^{2}}{1000} + \cos{\left((2.3059114360) \right)}} = 2.3058799487$ $LaTeX: x_{5} = (2.3058799487) - \frac{- \frac{713 (2.3058799487)^{3}}{1000} + \sin{\left((2.3058799487) \right)} + 8}{- \frac{2139 (2.3058799487)^{2}}{1000} + \cos{\left((2.3058799487) \right)}} = 2.3058799483$