Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{349 x^{3}}{500} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{349 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 3}{- \frac{1047 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{349 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 3}{- \frac{1047 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 1.9682610799$ $LaTeX: x_{2} = (1.9682610799) - \frac{- \frac{349 (1.9682610799)^{3}}{500} + \cos{\left((1.9682610799) \right)} + 3}{- \frac{1047 (1.9682610799)^{2}}{500} - \sin{\left((1.9682610799) \right)}} = 1.6683562251$ $LaTeX: x_{3} = (1.6683562251) - \frac{- \frac{349 (1.6683562251)^{3}}{500} + \cos{\left((1.6683562251) \right)} + 3}{- \frac{1047 (1.6683562251)^{2}}{500} - \sin{\left((1.6683562251) \right)}} = 1.6187169192$ $LaTeX: x_{4} = (1.6187169192) - \frac{- \frac{349 (1.6187169192)^{3}}{500} + \cos{\left((1.6187169192) \right)} + 3}{- \frac{1047 (1.6187169192)^{2}}{500} - \sin{\left((1.6187169192) \right)}} = 1.6174181709$ $LaTeX: x_{5} = (1.6174181709) - \frac{- \frac{349 (1.6174181709)^{3}}{500} + \cos{\left((1.6174181709) \right)} + 3}{- \frac{1047 (1.6174181709)^{2}}{500} - \sin{\left((1.6174181709) \right)}} = 1.6174172946$