Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{471 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{471 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 9}{- \frac{1413 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{471 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 9}{- \frac{1413 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.7391200154$ $LaTeX: x_{2} = (2.7391200154) - \frac{- \frac{471 (2.7391200154)^{3}}{1000} + \sin{\left((2.7391200154) \right)} + 9}{- \frac{1413 (2.7391200154)^{2}}{1000} + \cos{\left((2.7391200154) \right)}} = 2.7141379485$ $LaTeX: x_{3} = (2.7141379485) - \frac{- \frac{471 (2.7141379485)^{3}}{1000} + \sin{\left((2.7141379485) \right)} + 9}{- \frac{1413 (2.7141379485)^{2}}{1000} + \cos{\left((2.7141379485) \right)}} = 2.7139141834$ $LaTeX: x_{4} = (2.7139141834) - \frac{- \frac{471 (2.7139141834)^{3}}{1000} + \sin{\left((2.7139141834) \right)} + 9}{- \frac{1413 (2.7139141834)^{2}}{1000} + \cos{\left((2.7139141834) \right)}} = 2.7139141655$ $LaTeX: x_{5} = (2.7139141655) - \frac{- \frac{471 (2.7139141655)^{3}}{1000} + \sin{\left((2.7139141655) \right)} + 9}{- \frac{1413 (2.7139141655)^{2}}{1000} + \cos{\left((2.7139141655) \right)}} = 2.7139141655$