Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{177 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{177 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 2}{- \frac{531 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{177 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 2}{- \frac{531 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.5427485833$ $LaTeX: x_{2} = (2.5427485833) - \frac{- \frac{177 (2.5427485833)^{3}}{1000} + \sin{\left((2.5427485833) \right)} + 2}{- \frac{531 (2.5427485833)^{2}}{1000} + \cos{\left((2.5427485833) \right)}} = 2.4614544215$ $LaTeX: x_{3} = (2.4614544215) - \frac{- \frac{177 (2.4614544215)^{3}}{1000} + \sin{\left((2.4614544215) \right)} + 2}{- \frac{531 (2.4614544215)^{2}}{1000} + \cos{\left((2.4614544215) \right)}} = 2.4587599496$ $LaTeX: x_{4} = (2.4587599496) - \frac{- \frac{177 (2.4587599496)^{3}}{1000} + \sin{\left((2.4587599496) \right)} + 2}{- \frac{531 (2.4587599496)^{2}}{1000} + \cos{\left((2.4587599496) \right)}} = 2.4587569964$ $LaTeX: x_{5} = (2.4587569964) - \frac{- \frac{177 (2.4587569964)^{3}}{1000} + \sin{\left((2.4587569964) \right)} + 2}{- \frac{531 (2.4587569964)^{2}}{1000} + \cos{\left((2.4587569964) \right)}} = 2.4587569964$