Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{739 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{739 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{2217 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{739 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{2217 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.2619944115$ $LaTeX: x_{2} = (1.2619944115) - \frac{- \frac{739 (1.2619944115)^{3}}{1000} + \cos{\left((1.2619944115) \right)} + 1}{- \frac{2217 (1.2619944115)^{2}}{1000} - \sin{\left((1.2619944115) \right)}} = 1.2215374231$ $LaTeX: x_{3} = (1.2215374231) - \frac{- \frac{739 (1.2215374231)^{3}}{1000} + \cos{\left((1.2215374231) \right)} + 1}{- \frac{2217 (1.2215374231)^{2}}{1000} - \sin{\left((1.2215374231) \right)}} = 1.2204098371$ $LaTeX: x_{4} = (1.2204098371) - \frac{- \frac{739 (1.2204098371)^{3}}{1000} + \cos{\left((1.2204098371) \right)} + 1}{- \frac{2217 (1.2204098371)^{2}}{1000} - \sin{\left((1.2204098371) \right)}} = 1.2204089741$ $LaTeX: x_{5} = (1.2204089741) - \frac{- \frac{739 (1.2204089741)^{3}}{1000} + \cos{\left((1.2204089741) \right)} + 1}{- \frac{2217 (1.2204089741)^{2}}{1000} - \sin{\left((1.2204089741) \right)}} = 1.2204089741$