Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{591 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{591 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{1773 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{591 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{1773 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.3630953686$ $LaTeX: x_{2} = (1.3630953686) - \frac{- \frac{591 (1.3630953686)^{3}}{1000} + \cos{\left((1.3630953686) \right)} + 1}{- \frac{1773 (1.3630953686)^{2}}{1000} - \sin{\left((1.3630953686) \right)}} = 1.2950842176$ $LaTeX: x_{3} = (1.2950842176) - \frac{- \frac{591 (1.2950842176)^{3}}{1000} + \cos{\left((1.2950842176) \right)} + 1}{- \frac{1773 (1.2950842176)^{2}}{1000} - \sin{\left((1.2950842176) \right)}} = 1.2921571463$ $LaTeX: x_{4} = (1.2921571463) - \frac{- \frac{591 (1.2921571463)^{3}}{1000} + \cos{\left((1.2921571463) \right)} + 1}{- \frac{1773 (1.2921571463)^{2}}{1000} - \sin{\left((1.2921571463) \right)}} = 1.2921518353$ $LaTeX: x_{5} = (1.2921518353) - \frac{- \frac{591 (1.2921518353)^{3}}{1000} + \cos{\left((1.2921518353) \right)} + 1}{- \frac{1773 (1.2921518353)^{2}}{1000} - \sin{\left((1.2921518353) \right)}} = 1.2921518353$