Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{199 x^{3}}{250} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{199 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 5}{- \frac{597 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{199 (1.0000000000)^{3}}{250} + \cos{\left((1.0000000000) \right)} + 5}{- \frac{597 (1.0000000000)^{2}}{250} - \sin{\left((1.0000000000) \right)}} = 2.4690648494$ $LaTeX: x_{2} = (2.4690648494) - \frac{- \frac{199 (2.4690648494)^{3}}{250} + \cos{\left((2.4690648494) \right)} + 5}{- \frac{597 (2.4690648494)^{2}}{250} - \sin{\left((2.4690648494) \right)}} = 1.9576498710$ $LaTeX: x_{3} = (1.9576498710) - \frac{- \frac{199 (1.9576498710)^{3}}{250} + \cos{\left((1.9576498710) \right)} + 5}{- \frac{597 (1.9576498710)^{2}}{250} - \sin{\left((1.9576498710) \right)}} = 1.8237668829$ $LaTeX: x_{4} = (1.8237668829) - \frac{- \frac{199 (1.8237668829)^{3}}{250} + \cos{\left((1.8237668829) \right)} + 5}{- \frac{597 (1.8237668829)^{2}}{250} - \sin{\left((1.8237668829) \right)}} = 1.8149149762$ $LaTeX: x_{5} = (1.8149149762) - \frac{- \frac{199 (1.8149149762)^{3}}{250} + \cos{\left((1.8149149762) \right)} + 5}{- \frac{597 (1.8149149762)^{2}}{250} - \sin{\left((1.8149149762) \right)}} = 1.8148775158$