Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{19 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=7$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{19 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{57 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 7$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (7.0000000000) - \frac{- \frac{19 (7.0000000000)^{3}}{1000} + \sin{\left((7.0000000000) \right)} + 8}{- \frac{57 (7.0000000000)^{2}}{1000} + \cos{\left((7.0000000000) \right)}} = 8.0494772030$ $LaTeX: x_{2} = (8.0494772030) - \frac{- \frac{19 (8.0494772030)^{3}}{1000} + \sin{\left((8.0494772030) \right)} + 8}{- \frac{57 (8.0494772030)^{2}}{1000} + \cos{\left((8.0494772030) \right)}} = 7.8105945761$ $LaTeX: x_{3} = (7.8105945761) - \frac{- \frac{19 (7.8105945761)^{3}}{1000} + \sin{\left((7.8105945761) \right)} + 8}{- \frac{57 (7.8105945761)^{2}}{1000} + \cos{\left((7.8105945761) \right)}} = 7.7948051865$ $LaTeX: x_{4} = (7.7948051865) - \frac{- \frac{19 (7.7948051865)^{3}}{1000} + \sin{\left((7.7948051865) \right)} + 8}{- \frac{57 (7.7948051865)^{2}}{1000} + \cos{\left((7.7948051865) \right)}} = 7.7947360290$ $LaTeX: x_{5} = (7.7947360290) - \frac{- \frac{19 (7.7947360290)^{3}}{1000} + \sin{\left((7.7947360290) \right)} + 8}{- \frac{57 (7.7947360290)^{2}}{1000} + \cos{\left((7.7947360290) \right)}} = 7.7947360276$