Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{683 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{683 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 5}{- \frac{2049 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{683 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 5}{- \frac{2049 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.6804535771$ $LaTeX: x_{2} = (2.6804535771) - \frac{- \frac{683 (2.6804535771)^{3}}{1000} + \cos{\left((2.6804535771) \right)} + 5}{- \frac{2049 (2.6804535771)^{2}}{1000} - \sin{\left((2.6804535771) \right)}} = 2.0838055288$ $LaTeX: x_{3} = (2.0838055288) - \frac{- \frac{683 (2.0838055288)^{3}}{1000} + \cos{\left((2.0838055288) \right)} + 5}{- \frac{2049 (2.0838055288)^{2}}{1000} - \sin{\left((2.0838055288) \right)}} = 1.9127608837$ $LaTeX: x_{4} = (1.9127608837) - \frac{- \frac{683 (1.9127608837)^{3}}{1000} + \cos{\left((1.9127608837) \right)} + 5}{- \frac{2049 (1.9127608837)^{2}}{1000} - \sin{\left((1.9127608837) \right)}} = 1.8991258406$ $LaTeX: x_{5} = (1.8991258406) - \frac{- \frac{683 (1.8991258406)^{3}}{1000} + \cos{\left((1.8991258406) \right)} + 5}{- \frac{2049 (1.8991258406)^{2}}{1000} - \sin{\left((1.8991258406) \right)}} = 1.8990423372$