Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{19 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{19 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{57 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{19 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 1}{- \frac{57 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 3.4179129367$ $LaTeX: x_{2} = (3.4179129367) - \frac{- \frac{19 (3.4179129367)^{3}}{1000} + \sin{\left((3.4179129367) \right)} + 1}{- \frac{57 (3.4179129367)^{2}}{1000} + \cos{\left((3.4179129367) \right)}} = 3.3985886855$ $LaTeX: x_{3} = (3.3985886855) - \frac{- \frac{19 (3.3985886855)^{3}}{1000} + \sin{\left((3.3985886855) \right)} + 1}{- \frac{57 (3.3985886855)^{2}}{1000} + \cos{\left((3.3985886855) \right)}} = 3.3985746382$ $LaTeX: x_{4} = (3.3985746382) - \frac{- \frac{19 (3.3985746382)^{3}}{1000} + \sin{\left((3.3985746382) \right)} + 1}{- \frac{57 (3.3985746382)^{2}}{1000} + \cos{\left((3.3985746382) \right)}} = 3.3985746382$ $LaTeX: x_{5} = (3.3985746382) - \frac{- \frac{19 (3.3985746382)^{3}}{1000} + \sin{\left((3.3985746382) \right)} + 1}{- \frac{57 (3.3985746382)^{2}}{1000} + \cos{\left((3.3985746382) \right)}} = 3.3985746382$