Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{7 x^{3}}{10} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{7 x_{n}^{3}}{10} + \cos{\left(x_{n} \right)} + 5}{- \frac{21 x_{n}^{2}}{10} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{7 (1.0000000000)^{3}}{10} + \cos{\left((1.0000000000) \right)} + 5}{- \frac{21 (1.0000000000)^{2}}{10} - \sin{\left((1.0000000000) \right)}} = 2.6455380083$ $LaTeX: x_{2} = (2.6455380083) - \frac{- \frac{7 (2.6455380083)^{3}}{10} + \cos{\left((2.6455380083) \right)} + 5}{- \frac{21 (2.6455380083)^{2}}{10} - \sin{\left((2.6455380083) \right)}} = 2.0629129490$ $LaTeX: x_{3} = (2.0629129490) - \frac{- \frac{7 (2.0629129490)^{3}}{10} + \cos{\left((2.0629129490) \right)} + 5}{- \frac{21 (2.0629129490)^{2}}{10} - \sin{\left((2.0629129490) \right)}} = 1.8981400597$ $LaTeX: x_{4} = (1.8981400597) - \frac{- \frac{7 (1.8981400597)^{3}}{10} + \cos{\left((1.8981400597) \right)} + 5}{- \frac{21 (1.8981400597)^{2}}{10} - \sin{\left((1.8981400597) \right)}} = 1.8853664533$ $LaTeX: x_{5} = (1.8853664533) - \frac{- \frac{7 (1.8853664533)^{3}}{10} + \cos{\left((1.8853664533) \right)} + 5}{- \frac{21 (1.8853664533)^{2}}{10} - \sin{\left((1.8853664533) \right)}} = 1.8852924205$