Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{127 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{127 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 7}{- \frac{381 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{127 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{381 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.8788429837$ $LaTeX: x_{2} = (2.8788429837) - \frac{- \frac{127 (2.8788429837)^{3}}{500} + \cos{\left((2.8788429837) \right)} + 7}{- \frac{381 (2.8788429837)^{2}}{500} - \sin{\left((2.8788429837) \right)}} = 2.8749054679$ $LaTeX: x_{3} = (2.8749054679) - \frac{- \frac{127 (2.8749054679)^{3}}{500} + \cos{\left((2.8749054679) \right)} + 7}{- \frac{381 (2.8749054679)^{2}}{500} - \sin{\left((2.8749054679) \right)}} = 2.8749014274$ $LaTeX: x_{4} = (2.8749014274) - \frac{- \frac{127 (2.8749014274)^{3}}{500} + \cos{\left((2.8749014274) \right)} + 7}{- \frac{381 (2.8749014274)^{2}}{500} - \sin{\left((2.8749014274) \right)}} = 2.8749014273$ $LaTeX: x_{5} = (2.8749014273) - \frac{- \frac{127 (2.8749014273)^{3}}{500} + \cos{\left((2.8749014273) \right)} + 7}{- \frac{381 (2.8749014273)^{2}}{500} - \sin{\left((2.8749014273) \right)}} = 2.8749014273$