Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{57 x^{3}}{500} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{57 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 6}{- \frac{171 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{57 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 6}{- \frac{171 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 3.6001663494$ $LaTeX: x_{2} = (3.6001663494) - \frac{- \frac{57 (3.6001663494)^{3}}{500} + \cos{\left((3.6001663494) \right)} + 6}{- \frac{171 (3.6001663494)^{2}}{500} - \sin{\left((3.6001663494) \right)}} = 3.5459801622$ $LaTeX: x_{3} = (3.5459801622) - \frac{- \frac{57 (3.5459801622)^{3}}{500} + \cos{\left((3.5459801622) \right)} + 6}{- \frac{171 (3.5459801622)^{2}}{500} - \sin{\left((3.5459801622) \right)}} = 3.5453993360$ $LaTeX: x_{4} = (3.5453993360) - \frac{- \frac{57 (3.5453993360)^{3}}{500} + \cos{\left((3.5453993360) \right)} + 6}{- \frac{171 (3.5453993360)^{2}}{500} - \sin{\left((3.5453993360) \right)}} = 3.5453992710$ $LaTeX: x_{5} = (3.5453992710) - \frac{- \frac{57 (3.5453992710)^{3}}{500} + \cos{\left((3.5453992710) \right)} + 6}{- \frac{171 (3.5453992710)^{2}}{500} - \sin{\left((3.5453992710) \right)}} = 3.5453992710$