Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{61 x^{3}}{100} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{61 x_{n}^{3}}{100} + \cos{\left(x_{n} \right)} + 1}{- \frac{183 x_{n}^{2}}{100} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{61 (1.0000000000)^{3}}{100} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{183 (1.0000000000)^{2}}{100} - \sin{\left((1.0000000000) \right)}} = 1.3482359760$ $LaTeX: x_{2} = (1.3482359760) - \frac{- \frac{61 (1.3482359760)^{3}}{100} + \cos{\left((1.3482359760) \right)} + 1}{- \frac{183 (1.3482359760)^{2}}{100} - \sin{\left((1.3482359760) \right)}} = 1.2844892768$ $LaTeX: x_{3} = (1.2844892768) - \frac{- \frac{61 (1.2844892768)^{3}}{100} + \cos{\left((1.2844892768) \right)} + 1}{- \frac{183 (1.2844892768)^{2}}{100} - \sin{\left((1.2844892768) \right)}} = 1.2818857399$ $LaTeX: x_{4} = (1.2818857399) - \frac{- \frac{61 (1.2818857399)^{3}}{100} + \cos{\left((1.2818857399) \right)} + 1}{- \frac{183 (1.2818857399)^{2}}{100} - \sin{\left((1.2818857399) \right)}} = 1.2818814827$ $LaTeX: x_{5} = (1.2818814827) - \frac{- \frac{61 (1.2818814827)^{3}}{100} + \cos{\left((1.2818814827) \right)} + 1}{- \frac{183 (1.2818814827)^{2}}{100} - \sin{\left((1.2818814827) \right)}} = 1.2818814827$