Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{183 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{183 x_{n}^{3}}{250} + \cos{\left(x_{n} \right)} + 6}{- \frac{549 x_{n}^{2}}{250} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{183 (1.0000000000)^{3}}{250} + \cos{\left((1.0000000000) \right)} + 6}{- \frac{549 (1.0000000000)^{2}}{250} - \sin{\left((1.0000000000) \right)}} = 2.9122165561$ $LaTeX: x_{2} = (2.9122165561) - \frac{- \frac{183 (2.9122165561)^{3}}{250} + \cos{\left((2.9122165561) \right)} + 6}{- \frac{549 (2.9122165561)^{2}}{250} - \sin{\left((2.9122165561) \right)}} = 2.2198038107$ $LaTeX: x_{3} = (2.2198038107) - \frac{- \frac{183 (2.2198038107)^{3}}{250} + \cos{\left((2.2198038107) \right)} + 6}{- \frac{549 (2.2198038107)^{2}}{250} - \sin{\left((2.2198038107) \right)}} = 1.9950470617$ $LaTeX: x_{4} = (1.9950470617) - \frac{- \frac{183 (1.9950470617)^{3}}{250} + \cos{\left((1.9950470617) \right)} + 6}{- \frac{549 (1.9950470617)^{2}}{250} - \sin{\left((1.9950470617) \right)}} = 1.9718144107$ $LaTeX: x_{5} = (1.9718144107) - \frac{- \frac{183 (1.9718144107)^{3}}{250} + \cos{\left((1.9718144107) \right)} + 6}{- \frac{549 (1.9718144107)^{2}}{250} - \sin{\left((1.9718144107) \right)}} = 1.9715769208$