Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{239 x^{3}}{250} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{239 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 4}{- \frac{717 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{239 (1.0000000000)^{3}}{250} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{717 (1.0000000000)^{2}}{250} - \sin{\left((1.0000000000) \right)}} = 1.9662570001$ $LaTeX: x_{2} = (1.9662570001) - \frac{- \frac{239 (1.9662570001)^{3}}{250} + \cos{\left((1.9662570001) \right)} + 4}{- \frac{717 (1.9662570001)^{2}}{250} - \sin{\left((1.9662570001) \right)}} = 1.6621497280$ $LaTeX: x_{3} = (1.6621497280) - \frac{- \frac{239 (1.6621497280)^{3}}{250} + \cos{\left((1.6621497280) \right)} + 4}{- \frac{717 (1.6621497280)^{2}}{250} - \sin{\left((1.6621497280) \right)}} = 1.6081924840$ $LaTeX: x_{4} = (1.6081924840) - \frac{- \frac{239 (1.6081924840)^{3}}{250} + \cos{\left((1.6081924840) \right)} + 4}{- \frac{717 (1.6081924840)^{2}}{250} - \sin{\left((1.6081924840) \right)}} = 1.6065740673$ $LaTeX: x_{5} = (1.6065740673) - \frac{- \frac{239 (1.6065740673)^{3}}{250} + \cos{\left((1.6065740673) \right)} + 4}{- \frac{717 (1.6065740673)^{2}}{250} - \sin{\left((1.6065740673) \right)}} = 1.6065726357$