Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{121 x^{3}}{200} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{121 x_{n}^{3}}{200} + \cos{\left(x_{n} \right)} + 4}{- \frac{363 x_{n}^{2}}{200} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{121 (1.0000000000)^{3}}{200} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{363 (1.0000000000)^{2}}{200} - \sin{\left((1.0000000000) \right)}} = 2.4814023298$ $LaTeX: x_{2} = (2.4814023298) - \frac{- \frac{121 (2.4814023298)^{3}}{200} + \cos{\left((2.4814023298) \right)} + 4}{- \frac{363 (2.4814023298)^{2}}{200} - \sin{\left((2.4814023298) \right)}} = 1.9695977080$ $LaTeX: x_{3} = (1.9695977080) - \frac{- \frac{121 (1.9695977080)^{3}}{200} + \cos{\left((1.9695977080) \right)} + 4}{- \frac{363 (1.9695977080)^{2}}{200} - \sin{\left((1.9695977080) \right)}} = 1.8426358680$ $LaTeX: x_{4} = (1.8426358680) - \frac{- \frac{121 (1.8426358680)^{3}}{200} + \cos{\left((1.8426358680) \right)} + 4}{- \frac{363 (1.8426358680)^{2}}{200} - \sin{\left((1.8426358680) \right)}} = 1.8351175037$ $LaTeX: x_{5} = (1.8351175037) - \frac{- \frac{121 (1.8351175037)^{3}}{200} + \cos{\left((1.8351175037) \right)} + 4}{- \frac{363 (1.8351175037)^{2}}{200} - \sin{\left((1.8351175037) \right)}} = 1.8350918923$