Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{34 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{34 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 7}{- \frac{102 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{34 (3.0000000000)^{3}}{125} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{102 (3.0000000000)^{2}}{125} - \sin{\left((3.0000000000) \right)}} = 2.8217807470$ $LaTeX: x_{2} = (2.8217807470) - \frac{- \frac{34 (2.8217807470)^{3}}{125} + \cos{\left((2.8217807470) \right)} + 7}{- \frac{102 (2.8217807470)^{2}}{125} - \sin{\left((2.8217807470) \right)}} = 2.8128746123$ $LaTeX: x_{3} = (2.8128746123) - \frac{- \frac{34 (2.8128746123)^{3}}{125} + \cos{\left((2.8128746123) \right)} + 7}{- \frac{102 (2.8128746123)^{2}}{125} - \sin{\left((2.8128746123) \right)}} = 2.8128532479$ $LaTeX: x_{4} = (2.8128532479) - \frac{- \frac{34 (2.8128532479)^{3}}{125} + \cos{\left((2.8128532479) \right)} + 7}{- \frac{102 (2.8128532479)^{2}}{125} - \sin{\left((2.8128532479) \right)}} = 2.8128532478$ $LaTeX: x_{5} = (2.8128532478) - \frac{- \frac{34 (2.8128532478)^{3}}{125} + \cos{\left((2.8128532478) \right)} + 7}{- \frac{102 (2.8128532478)^{2}}{125} - \sin{\left((2.8128532478) \right)}} = 2.8128532478$