Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{18 x^{3}}{125} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{18 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 2}{- \frac{54 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{18 (3.0000000000)^{3}}{125} + \cos{\left((3.0000000000) \right)} + 2}{- \frac{54 (3.0000000000)^{2}}{125} - \sin{\left((3.0000000000) \right)}} = 2.2857019670$ $LaTeX: x_{2} = (2.2857019670) - \frac{- \frac{18 (2.2857019670)^{3}}{125} + \cos{\left((2.2857019670) \right)} + 2}{- \frac{54 (2.2857019670)^{2}}{125} - \sin{\left((2.2857019670) \right)}} = 2.1611640805$ $LaTeX: x_{3} = (2.1611640805) - \frac{- \frac{18 (2.1611640805)^{3}}{125} + \cos{\left((2.1611640805) \right)} + 2}{- \frac{54 (2.1611640805)^{2}}{125} - \sin{\left((2.1611640805) \right)}} = 2.1575823741$ $LaTeX: x_{4} = (2.1575823741) - \frac{- \frac{18 (2.1575823741)^{3}}{125} + \cos{\left((2.1575823741) \right)} + 2}{- \frac{54 (2.1575823741)^{2}}{125} - \sin{\left((2.1575823741) \right)}} = 2.1575794181$ $LaTeX: x_{5} = (2.1575794181) - \frac{- \frac{18 (2.1575794181)^{3}}{125} + \cos{\left((2.1575794181) \right)} + 2}{- \frac{54 (2.1575794181)^{2}}{125} - \sin{\left((2.1575794181) \right)}} = 2.1575794181$