Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{587 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{587 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 4}{- \frac{1761 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{587 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{1761 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.5190572071$ $LaTeX: x_{2} = (2.5190572071) - \frac{- \frac{587 (2.5190572071)^{3}}{1000} + \cos{\left((2.5190572071) \right)} + 4}{- \frac{1761 (2.5190572071)^{2}}{1000} - \sin{\left((2.5190572071) \right)}} = 1.9921187623$ $LaTeX: x_{3} = (1.9921187623) - \frac{- \frac{587 (1.9921187623)^{3}}{1000} + \cos{\left((1.9921187623) \right)} + 4}{- \frac{1761 (1.9921187623)^{2}}{1000} - \sin{\left((1.9921187623) \right)}} = 1.8592682802$ $LaTeX: x_{4} = (1.8592682802) - \frac{- \frac{587 (1.8592682802)^{3}}{1000} + \cos{\left((1.8592682802) \right)} + 4}{- \frac{1761 (1.8592682802)^{2}}{1000} - \sin{\left((1.8592682802) \right)}} = 1.8511374252$ $LaTeX: x_{5} = (1.8511374252) - \frac{- \frac{587 (1.8511374252)^{3}}{1000} + \cos{\left((1.8511374252) \right)} + 4}{- \frac{1761 (1.8511374252)^{2}}{1000} - \sin{\left((1.8511374252) \right)}} = 1.8511078593$