Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{179 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{179 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{537 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{179 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 1}{- \frac{537 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3659823546$ $LaTeX: x_{2} = (2.3659823546) - \frac{- \frac{179 (2.3659823546)^{3}}{1000} + \sin{\left((2.3659823546) \right)} + 1}{- \frac{537 (2.3659823546)^{2}}{1000} + \cos{\left((2.3659823546) \right)}} = 2.1857139479$ $LaTeX: x_{3} = (2.1857139479) - \frac{- \frac{179 (2.1857139479)^{3}}{1000} + \sin{\left((2.1857139479) \right)} + 1}{- \frac{537 (2.1857139479)^{2}}{1000} + \cos{\left((2.1857139479) \right)}} = 2.1690762629$ $LaTeX: x_{4} = (2.1690762629) - \frac{- \frac{179 (2.1690762629)^{3}}{1000} + \sin{\left((2.1690762629) \right)} + 1}{- \frac{537 (2.1690762629)^{2}}{1000} + \cos{\left((2.1690762629) \right)}} = 2.1689346424$ $LaTeX: x_{5} = (2.1689346424) - \frac{- \frac{179 (2.1689346424)^{3}}{1000} + \sin{\left((2.1689346424) \right)} + 1}{- \frac{537 (2.1689346424)^{2}}{1000} + \cos{\left((2.1689346424) \right)}} = 2.1689346322$