Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{47 x^{3}}{200} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{47 x_{n}^{3}}{200} + \sin{\left(x_{n} \right)} + 6}{- \frac{141 x_{n}^{2}}{200} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{47 (3.0000000000)^{3}}{200} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{141 (3.0000000000)^{2}}{200} + \cos{\left((3.0000000000) \right)}} = 2.9722044716$ $LaTeX: x_{2} = (2.9722044716) - \frac{- \frac{47 (2.9722044716)^{3}}{200} + \sin{\left((2.9722044716) \right)} + 6}{- \frac{141 (2.9722044716)^{2}}{200} + \cos{\left((2.9722044716) \right)}} = 2.9719706044$ $LaTeX: x_{3} = (2.9719706044) - \frac{- \frac{47 (2.9719706044)^{3}}{200} + \sin{\left((2.9719706044) \right)} + 6}{- \frac{141 (2.9719706044)^{2}}{200} + \cos{\left((2.9719706044) \right)}} = 2.9719705879$ $LaTeX: x_{4} = (2.9719705879) - \frac{- \frac{47 (2.9719705879)^{3}}{200} + \sin{\left((2.9719705879) \right)} + 6}{- \frac{141 (2.9719705879)^{2}}{200} + \cos{\left((2.9719705879) \right)}} = 2.9719705879$ $LaTeX: x_{5} = (2.9719705879) - \frac{- \frac{47 (2.9719705879)^{3}}{200} + \sin{\left((2.9719705879) \right)} + 6}{- \frac{141 (2.9719705879)^{2}}{200} + \cos{\left((2.9719705879) \right)}} = 2.9719705879$