Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{104 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{104 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 7}{- \frac{312 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{104 (1.0000000000)^{3}}{125} + \cos{\left((1.0000000000) \right)} + 7}{- \frac{312 (1.0000000000)^{2}}{125} - \sin{\left((1.0000000000) \right)}} = 3.0099956933$ $LaTeX: x_{2} = (3.0099956933) - \frac{- \frac{104 (3.0099956933)^{3}}{125} + \cos{\left((3.0099956933) \right)} + 7}{- \frac{312 (3.0099956933)^{2}}{125} - \sin{\left((3.0099956933) \right)}} = 2.2766245442$ $LaTeX: x_{3} = (2.2766245442) - \frac{- \frac{104 (2.2766245442)^{3}}{125} + \cos{\left((2.2766245442) \right)} + 7}{- \frac{312 (2.2766245442)^{2}}{125} - \sin{\left((2.2766245442) \right)}} = 2.0235864030$ $LaTeX: x_{4} = (2.0235864030) - \frac{- \frac{104 (2.0235864030)^{3}}{125} + \cos{\left((2.0235864030) \right)} + 7}{- \frac{312 (2.0235864030)^{2}}{125} - \sin{\left((2.0235864030) \right)}} = 1.9937529749$ $LaTeX: x_{5} = (1.9937529749) - \frac{- \frac{104 (1.9937529749)^{3}}{125} + \cos{\left((1.9937529749) \right)} + 7}{- \frac{312 (1.9937529749)^{2}}{125} - \sin{\left((1.9937529749) \right)}} = 1.9933576622$