Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{13 x^{3}}{20} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{13 x_{n}^{3}}{20} + \cos{\left(x_{n} \right)} + 4}{- \frac{39 x_{n}^{2}}{20} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{13 (1.0000000000)^{3}}{20} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{39 (1.0000000000)^{2}}{20} - \sin{\left((1.0000000000) \right)}} = 2.3936388116$ $LaTeX: x_{2} = (2.3936388116) - \frac{- \frac{13 (2.3936388116)^{3}}{20} + \cos{\left((2.3936388116) \right)} + 4}{- \frac{39 (2.3936388116)^{2}}{20} - \sin{\left((2.3936388116) \right)}} = 1.9171707727$ $LaTeX: x_{3} = (1.9171707727) - \frac{- \frac{13 (1.9171707727)^{3}}{20} + \cos{\left((1.9171707727) \right)} + 4}{- \frac{39 (1.9171707727)^{2}}{20} - \sin{\left((1.9171707727) \right)}} = 1.8037250165$ $LaTeX: x_{4} = (1.8037250165) - \frac{- \frac{13 (1.8037250165)^{3}}{20} + \cos{\left((1.8037250165) \right)} + 4}{- \frac{39 (1.8037250165)^{2}}{20} - \sin{\left((1.8037250165) \right)}} = 1.7975461943$ $LaTeX: x_{5} = (1.7975461943) - \frac{- \frac{13 (1.7975461943)^{3}}{20} + \cos{\left((1.7975461943) \right)} + 4}{- \frac{39 (1.7975461943)^{2}}{20} - \sin{\left((1.7975461943) \right)}} = 1.7975283583$