Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{189 x^{3}}{200} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{189 x_{n}^{3}}{200} + \cos{\left(x_{n} \right)} + 5}{- \frac{567 x_{n}^{2}}{200} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{189 (1.0000000000)^{3}}{200} + \cos{\left((1.0000000000) \right)} + 5}{- \frac{567 (1.0000000000)^{2}}{200} - \sin{\left((1.0000000000) \right)}} = 2.2499220924$ $LaTeX: x_{2} = (2.2499220924) - \frac{- \frac{189 (2.2499220924)^{3}}{200} + \cos{\left((2.2499220924) \right)} + 5}{- \frac{567 (2.2499220924)^{2}}{200} - \sin{\left((2.2499220924) \right)}} = 1.8274882144$ $LaTeX: x_{3} = (1.8274882144) - \frac{- \frac{189 (1.8274882144)^{3}}{200} + \cos{\left((1.8274882144) \right)} + 5}{- \frac{567 (1.8274882144)^{2}}{200} - \sin{\left((1.8274882144) \right)}} = 1.7296006806$ $LaTeX: x_{4} = (1.7296006806) - \frac{- \frac{189 (1.7296006806)^{3}}{200} + \cos{\left((1.7296006806) \right)} + 5}{- \frac{567 (1.7296006806)^{2}}{200} - \sin{\left((1.7296006806) \right)}} = 1.7245636065$ $LaTeX: x_{5} = (1.7245636065) - \frac{- \frac{189 (1.7245636065)^{3}}{200} + \cos{\left((1.7245636065) \right)} + 5}{- \frac{567 (1.7245636065)^{2}}{200} - \sin{\left((1.7245636065) \right)}} = 1.7245506228$