Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{587 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{587 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 5}{- \frac{1761 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{587 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{1761 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3641020985$ $LaTeX: x_{2} = (2.3641020985) - \frac{- \frac{587 (2.3641020985)^{3}}{1000} + \sin{\left((2.3641020985) \right)} + 5}{- \frac{1761 (2.3641020985)^{2}}{1000} + \cos{\left((2.3641020985) \right)}} = 2.1694536434$ $LaTeX: x_{3} = (2.1694536434) - \frac{- \frac{587 (2.1694536434)^{3}}{1000} + \sin{\left((2.1694536434) \right)} + 5}{- \frac{1761 (2.1694536434)^{2}}{1000} + \cos{\left((2.1694536434) \right)}} = 2.1505277111$ $LaTeX: x_{4} = (2.1505277111) - \frac{- \frac{587 (2.1505277111)^{3}}{1000} + \sin{\left((2.1505277111) \right)} + 5}{- \frac{1761 (2.1505277111)^{2}}{1000} + \cos{\left((2.1505277111) \right)}} = 2.1503536390$ $LaTeX: x_{5} = (2.1503536390) - \frac{- \frac{587 (2.1503536390)^{3}}{1000} + \sin{\left((2.1503536390) \right)} + 5}{- \frac{1761 (2.1503536390)^{2}}{1000} + \cos{\left((2.1503536390) \right)}} = 2.1503536244$