Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{881 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{881 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 7}{- \frac{2643 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{881 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 7}{- \frac{2643 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.9111372529$ $LaTeX: x_{2} = (2.9111372529) - \frac{- \frac{881 (2.9111372529)^{3}}{1000} + \cos{\left((2.9111372529) \right)} + 7}{- \frac{2643 (2.9111372529)^{2}}{1000} - \sin{\left((2.9111372529) \right)}} = 2.2168912436$ $LaTeX: x_{3} = (2.2168912436) - \frac{- \frac{881 (2.2168912436)^{3}}{1000} + \cos{\left((2.2168912436) \right)} + 7}{- \frac{2643 (2.2168912436)^{2}}{1000} - \sin{\left((2.2168912436) \right)}} = 1.9847504066$ $LaTeX: x_{4} = (1.9847504066) - \frac{- \frac{881 (1.9847504066)^{3}}{1000} + \cos{\left((1.9847504066) \right)} + 7}{- \frac{2643 (1.9847504066)^{2}}{1000} - \sin{\left((1.9847504066) \right)}} = 1.9591265247$ $LaTeX: x_{5} = (1.9591265247) - \frac{- \frac{881 (1.9591265247)^{3}}{1000} + \cos{\left((1.9591265247) \right)} + 7}{- \frac{2643 (1.9591265247)^{2}}{1000} - \sin{\left((1.9591265247) \right)}} = 1.9588284230$