Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{313 x^{3}}{500} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{313 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 9}{- \frac{939 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{313 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{939 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.4782649836$ $LaTeX: x_{2} = (2.4782649836) - \frac{- \frac{313 (2.4782649836)^{3}}{500} + \cos{\left((2.4782649836) \right)} + 9}{- \frac{939 (2.4782649836)^{2}}{500} - \sin{\left((2.4782649836) \right)}} = 2.3699283320$ $LaTeX: x_{3} = (2.3699283320) - \frac{- \frac{313 (2.3699283320)^{3}}{500} + \cos{\left((2.3699283320) \right)} + 9}{- \frac{939 (2.3699283320)^{2}}{500} - \sin{\left((2.3699283320) \right)}} = 2.3655406677$ $LaTeX: x_{4} = (2.3655406677) - \frac{- \frac{313 (2.3655406677)^{3}}{500} + \cos{\left((2.3655406677) \right)} + 9}{- \frac{939 (2.3655406677)^{2}}{500} - \sin{\left((2.3655406677) \right)}} = 2.3655336431$ $LaTeX: x_{5} = (2.3655336431) - \frac{- \frac{313 (2.3655336431)^{3}}{500} + \cos{\left((2.3655336431) \right)} + 9}{- \frac{939 (2.3655336431)^{2}}{500} - \sin{\left((2.3655336431) \right)}} = 2.3655336431$