Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{851 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{851 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{2553 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{851 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{2553 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 1.4921111539$ $LaTeX: x_{2} = (1.4921111539) - \frac{- \frac{851 (1.4921111539)^{3}}{1000} + \sin{\left((1.4921111539) \right)} + 1}{- \frac{2553 (1.4921111539)^{2}}{1000} + \cos{\left((1.4921111539) \right)}} = 1.3440139935$ $LaTeX: x_{3} = (1.3440139935) - \frac{- \frac{851 (1.3440139935)^{3}}{1000} + \sin{\left((1.3440139935) \right)} + 1}{- \frac{2553 (1.3440139935)^{2}}{1000} + \cos{\left((1.3440139935) \right)}} = 1.3231206335$ $LaTeX: x_{4} = (1.3231206335) - \frac{- \frac{851 (1.3231206335)^{3}}{1000} + \sin{\left((1.3231206335) \right)} + 1}{- \frac{2553 (1.3231206335)^{2}}{1000} + \cos{\left((1.3231206335) \right)}} = 1.3227176212$ $LaTeX: x_{5} = (1.3227176212) - \frac{- \frac{851 (1.3227176212)^{3}}{1000} + \sin{\left((1.3227176212) \right)} + 1}{- \frac{2553 (1.3227176212)^{2}}{1000} + \cos{\left((1.3227176212) \right)}} = 1.3227174725$