Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{521 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{521 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 3}{- \frac{1563 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{521 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 3}{- \frac{1563 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.2557033647$ $LaTeX: x_{2} = (2.2557033647) - \frac{- \frac{521 (2.2557033647)^{3}}{1000} + \cos{\left((2.2557033647) \right)} + 3}{- \frac{1563 (2.2557033647)^{2}}{1000} - \sin{\left((2.2557033647) \right)}} = 1.8417898610$ $LaTeX: x_{3} = (1.8417898610) - \frac{- \frac{521 (1.8417898610)^{3}}{1000} + \cos{\left((1.8417898610) \right)} + 3}{- \frac{1563 (1.8417898610)^{2}}{1000} - \sin{\left((1.8417898610) \right)}} = 1.7583581815$ $LaTeX: x_{4} = (1.7583581815) - \frac{- \frac{521 (1.7583581815)^{3}}{1000} + \cos{\left((1.7583581815) \right)} + 3}{- \frac{1563 (1.7583581815)^{2}}{1000} - \sin{\left((1.7583581815) \right)}} = 1.7551083293$ $LaTeX: x_{5} = (1.7551083293) - \frac{- \frac{521 (1.7551083293)^{3}}{1000} + \cos{\left((1.7551083293) \right)} + 3}{- \frac{1563 (1.7551083293)^{2}}{1000} - \sin{\left((1.7551083293) \right)}} = 1.7551034948$