Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{67 x^{3}}{250} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{67 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 4}{- \frac{201 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{67 (3.0000000000)^{3}}{250} + \cos{\left((3.0000000000) \right)} + 4}{- \frac{201 (3.0000000000)^{2}}{250} - \sin{\left((3.0000000000) \right)}} = 2.4271487393$ $LaTeX: x_{2} = (2.4271487393) - \frac{- \frac{67 (2.4271487393)^{3}}{250} + \cos{\left((2.4271487393) \right)} + 4}{- \frac{201 (2.4271487393)^{2}}{250} - \sin{\left((2.4271487393) \right)}} = 2.3181933421$ $LaTeX: x_{3} = (2.3181933421) - \frac{- \frac{67 (2.3181933421)^{3}}{250} + \cos{\left((2.3181933421) \right)} + 4}{- \frac{201 (2.3181933421)^{2}}{250} - \sin{\left((2.3181933421) \right)}} = 2.3145368057$ $LaTeX: x_{4} = (2.3145368057) - \frac{- \frac{67 (2.3145368057)^{3}}{250} + \cos{\left((2.3145368057) \right)} + 4}{- \frac{201 (2.3145368057)^{2}}{250} - \sin{\left((2.3145368057) \right)}} = 2.3145327667$ $LaTeX: x_{5} = (2.3145327667) - \frac{- \frac{67 (2.3145327667)^{3}}{250} + \cos{\left((2.3145327667) \right)} + 4}{- \frac{201 (2.3145327667)^{2}}{250} - \sin{\left((2.3145327667) \right)}} = 2.3145327667$