Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{303 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{303 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 4}{- \frac{909 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{303 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{909 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 2.4793552283$ $LaTeX: x_{2} = (2.4793552283) - \frac{- \frac{303 (2.4793552283)^{3}}{500} + \cos{\left((2.4793552283) \right)} + 4}{- \frac{909 (2.4793552283)^{2}}{500} - \sin{\left((2.4793552283) \right)}} = 1.9683738721$ $LaTeX: x_{3} = (1.9683738721) - \frac{- \frac{303 (1.9683738721)^{3}}{500} + \cos{\left((1.9683738721) \right)} + 4}{- \frac{909 (1.9683738721)^{2}}{500} - \sin{\left((1.9683738721) \right)}} = 1.8417307273$ $LaTeX: x_{4} = (1.8417307273) - \frac{- \frac{303 (1.8417307273)^{3}}{500} + \cos{\left((1.8417307273) \right)} + 4}{- \frac{909 (1.8417307273)^{2}}{500} - \sin{\left((1.8417307273) \right)}} = 1.8342450287$ $LaTeX: x_{5} = (1.8342450287) - \frac{- \frac{303 (1.8342450287)^{3}}{500} + \cos{\left((1.8342450287) \right)} + 4}{- \frac{909 (1.8342450287)^{2}}{500} - \sin{\left((1.8342450287) \right)}} = 1.8342196213$