Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{741 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{741 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 2}{- \frac{2223 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{741 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{2223 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.5871494019$ $LaTeX: x_{2} = (1.5871494019) - \frac{- \frac{741 (1.5871494019)^{3}}{1000} + \cos{\left((1.5871494019) \right)} + 2}{- \frac{2223 (1.5871494019)^{2}}{1000} - \sin{\left((1.5871494019) \right)}} = 1.4388179601$ $LaTeX: x_{3} = (1.4388179601) - \frac{- \frac{741 (1.4388179601)^{3}}{1000} + \cos{\left((1.4388179601) \right)} + 2}{- \frac{2223 (1.4388179601)^{2}}{1000} - \sin{\left((1.4388179601) \right)}} = 1.4253065071$ $LaTeX: x_{4} = (1.4253065071) - \frac{- \frac{741 (1.4253065071)^{3}}{1000} + \cos{\left((1.4253065071) \right)} + 2}{- \frac{2223 (1.4253065071)^{2}}{1000} - \sin{\left((1.4253065071) \right)}} = 1.4251985222$ $LaTeX: x_{5} = (1.4251985222) - \frac{- \frac{741 (1.4251985222)^{3}}{1000} + \cos{\left((1.4251985222) \right)} + 2}{- \frac{2223 (1.4251985222)^{2}}{1000} - \sin{\left((1.4251985222) \right)}} = 1.4251985153$