Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{407 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{407 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 3}{- \frac{1221 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{407 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 3}{- \frac{1221 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3448631015$ $LaTeX: x_{2} = (2.3448631015) - \frac{- \frac{407 (2.3448631015)^{3}}{1000} + \sin{\left((2.3448631015) \right)} + 3}{- \frac{1221 (2.3448631015)^{2}}{1000} + \cos{\left((2.3448631015) \right)}} = 2.1381388288$ $LaTeX: x_{3} = (2.1381388288) - \frac{- \frac{407 (2.1381388288)^{3}}{1000} + \sin{\left((2.1381388288) \right)} + 3}{- \frac{1221 (2.1381388288)^{2}}{1000} + \cos{\left((2.1381388288) \right)}} = 2.1160760907$ $LaTeX: x_{4} = (2.1160760907) - \frac{- \frac{407 (2.1160760907)^{3}}{1000} + \sin{\left((2.1160760907) \right)} + 3}{- \frac{1221 (2.1160760907)^{2}}{1000} + \cos{\left((2.1160760907) \right)}} = 2.1158300819$ $LaTeX: x_{5} = (2.1158300819) - \frac{- \frac{407 (2.1158300819)^{3}}{1000} + \sin{\left((2.1158300819) \right)} + 3}{- \frac{1221 (2.1158300819)^{2}}{1000} + \cos{\left((2.1158300819) \right)}} = 2.1158300514$