Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{61 x^{3}}{200} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{61 x_{n}^{3}}{200} + \cos{\left(x_{n} \right)} + 9}{- \frac{183 x_{n}^{2}}{200} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{61 (3.0000000000)^{3}}{200} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{183 (3.0000000000)^{2}}{200} - \sin{\left((3.0000000000) \right)}} = 2.9731388165$ $LaTeX: x_{2} = (2.9731388165) - \frac{- \frac{61 (2.9731388165)^{3}}{200} + \cos{\left((2.9731388165) \right)} + 9}{- \frac{183 (2.9731388165)^{2}}{200} - \sin{\left((2.9731388165) \right)}} = 2.9729428348$ $LaTeX: x_{3} = (2.9729428348) - \frac{- \frac{61 (2.9729428348)^{3}}{200} + \cos{\left((2.9729428348) \right)} + 9}{- \frac{183 (2.9729428348)^{2}}{200} - \sin{\left((2.9729428348) \right)}} = 2.9729428244$ $LaTeX: x_{4} = (2.9729428244) - \frac{- \frac{61 (2.9729428244)^{3}}{200} + \cos{\left((2.9729428244) \right)} + 9}{- \frac{183 (2.9729428244)^{2}}{200} - \sin{\left((2.9729428244) \right)}} = 2.9729428244$ $LaTeX: x_{5} = (2.9729428244) - \frac{- \frac{61 (2.9729428244)^{3}}{200} + \cos{\left((2.9729428244) \right)} + 9}{- \frac{183 (2.9729428244)^{2}}{200} - \sin{\left((2.9729428244) \right)}} = 2.9729428244$