Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{201 x^{3}}{250} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{201 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 3}{- \frac{603 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{201 (1.0000000000)^{3}}{250} + \cos{\left((1.0000000000) \right)} + 3}{- \frac{603 (1.0000000000)^{2}}{250} - \sin{\left((1.0000000000) \right)}} = 1.8410409432$ $LaTeX: x_{2} = (1.8410409432) - \frac{- \frac{201 (1.8410409432)^{3}}{250} + \cos{\left((1.8410409432) \right)} + 3}{- \frac{603 (1.8410409432)^{2}}{250} - \sin{\left((1.8410409432) \right)}} = 1.5911241176$ $LaTeX: x_{3} = (1.5911241176) - \frac{- \frac{201 (1.5911241176)^{3}}{250} + \cos{\left((1.5911241176) \right)} + 3}{- \frac{603 (1.5911241176)^{2}}{250} - \sin{\left((1.5911241176) \right)}} = 1.5546759560$ $LaTeX: x_{4} = (1.5546759560) - \frac{- \frac{201 (1.5546759560)^{3}}{250} + \cos{\left((1.5546759560) \right)} + 3}{- \frac{603 (1.5546759560)^{2}}{250} - \sin{\left((1.5546759560) \right)}} = 1.5539359512$ $LaTeX: x_{5} = (1.5539359512) - \frac{- \frac{201 (1.5539359512)^{3}}{250} + \cos{\left((1.5539359512) \right)} + 3}{- \frac{603 (1.5539359512)^{2}}{250} - \sin{\left((1.5539359512) \right)}} = 1.5539356497$