Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{71 x^{3}}{200} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{71 x_{n}^{3}}{200} + \cos{\left(x_{n} \right)} + 6}{- \frac{213 x_{n}^{2}}{200} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{71 (3.0000000000)^{3}}{200} + \cos{\left((3.0000000000) \right)} + 6}{- \frac{213 (3.0000000000)^{2}}{200} - \sin{\left((3.0000000000) \right)}} = 2.5296179265$ $LaTeX: x_{2} = (2.5296179265) - \frac{- \frac{71 (2.5296179265)^{3}}{200} + \cos{\left((2.5296179265) \right)} + 6}{- \frac{213 (2.5296179265)^{2}}{200} - \sin{\left((2.5296179265) \right)}} = 2.4531732235$ $LaTeX: x_{3} = (2.4531732235) - \frac{- \frac{71 (2.4531732235)^{3}}{200} + \cos{\left((2.4531732235) \right)} + 6}{- \frac{213 (2.4531732235)^{2}}{200} - \sin{\left((2.4531732235) \right)}} = 2.4512941611$ $LaTeX: x_{4} = (2.4512941611) - \frac{- \frac{71 (2.4512941611)^{3}}{200} + \cos{\left((2.4512941611) \right)} + 6}{- \frac{213 (2.4512941611)^{2}}{200} - \sin{\left((2.4512941611) \right)}} = 2.4512930440$ $LaTeX: x_{5} = (2.4512930440) - \frac{- \frac{71 (2.4512930440)^{3}}{200} + \cos{\left((2.4512930440) \right)} + 6}{- \frac{213 (2.4512930440)^{2}}{200} - \sin{\left((2.4512930440) \right)}} = 2.4512930440$