Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{13 x^{3}}{125} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{13 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 2}{- \frac{39 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{13 (3.0000000000)^{3}}{125} + \cos{\left((3.0000000000) \right)} + 2}{- \frac{39 (3.0000000000)^{2}}{125} - \sin{\left((3.0000000000) \right)}} = 2.3903291519$ $LaTeX: x_{2} = (2.3903291519) - \frac{- \frac{13 (2.3903291519)^{3}}{125} + \cos{\left((2.3903291519) \right)} + 2}{- \frac{39 (2.3903291519)^{2}}{125} - \sin{\left((2.3903291519) \right)}} = 2.3289907094$ $LaTeX: x_{3} = (2.3289907094) - \frac{- \frac{13 (2.3289907094)^{3}}{125} + \cos{\left((2.3289907094) \right)} + 2}{- \frac{39 (2.3289907094)^{2}}{125} - \sin{\left((2.3289907094) \right)}} = 2.3283978528$ $LaTeX: x_{4} = (2.3283978528) - \frac{- \frac{13 (2.3283978528)^{3}}{125} + \cos{\left((2.3283978528) \right)} + 2}{- \frac{39 (2.3283978528)^{2}}{125} - \sin{\left((2.3283978528) \right)}} = 2.3283977971$ $LaTeX: x_{5} = (2.3283977971) - \frac{- \frac{13 (2.3283977971)^{3}}{125} + \cos{\left((2.3283977971) \right)} + 2}{- \frac{39 (2.3283977971)^{2}}{125} - \sin{\left((2.3283977971) \right)}} = 2.3283977971$