Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{691 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{691 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 2}{- \frac{2073 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{691 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{2073 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.6345241780$ $LaTeX: x_{2} = (1.6345241780) - \frac{- \frac{691 (1.6345241780)^{3}}{1000} + \cos{\left((1.6345241780) \right)} + 2}{- \frac{2073 (1.6345241780)^{2}}{1000} - \sin{\left((1.6345241780) \right)}} = 1.4691077367$ $LaTeX: x_{3} = (1.4691077367) - \frac{- \frac{691 (1.4691077367)^{3}}{1000} + \cos{\left((1.4691077367) \right)} + 2}{- \frac{2073 (1.4691077367)^{2}}{1000} - \sin{\left((1.4691077367) \right)}} = 1.4527481864$ $LaTeX: x_{4} = (1.4527481864) - \frac{- \frac{691 (1.4527481864)^{3}}{1000} + \cos{\left((1.4527481864) \right)} + 2}{- \frac{2073 (1.4527481864)^{2}}{1000} - \sin{\left((1.4527481864) \right)}} = 1.4525942470$ $LaTeX: x_{5} = (1.4525942470) - \frac{- \frac{691 (1.4525942470)^{3}}{1000} + \cos{\left((1.4525942470) \right)} + 2}{- \frac{2073 (1.4525942470)^{2}}{1000} - \sin{\left((1.4525942470) \right)}} = 1.4525942334$