Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{122 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{122 x_{n}^{3}}{125} + \cos{\left(x_{n} \right)} + 7}{- \frac{366 x_{n}^{2}}{125} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{122 (1.0000000000)^{3}}{125} + \cos{\left((1.0000000000) \right)} + 7}{- \frac{366 (1.0000000000)^{2}}{125} - \sin{\left((1.0000000000) \right)}} = 2.7414386083$ $LaTeX: x_{2} = (2.7414386083) - \frac{- \frac{122 (2.7414386083)^{3}}{125} + \cos{\left((2.7414386083) \right)} + 7}{- \frac{366 (2.7414386083)^{2}}{125} - \sin{\left((2.7414386083) \right)}} = 2.1149672105$ $LaTeX: x_{3} = (2.1149672105) - \frac{- \frac{122 (2.1149672105)^{3}}{125} + \cos{\left((2.1149672105) \right)} + 7}{- \frac{366 (2.1149672105)^{2}}{125} - \sin{\left((2.1149672105) \right)}} = 1.9177954174$ $LaTeX: x_{4} = (1.9177954174) - \frac{- \frac{122 (1.9177954174)^{3}}{125} + \cos{\left((1.9177954174) \right)} + 7}{- \frac{366 (1.9177954174)^{2}}{125} - \sin{\left((1.9177954174) \right)}} = 1.8986375231$ $LaTeX: x_{5} = (1.8986375231) - \frac{- \frac{122 (1.8986375231)^{3}}{125} + \cos{\left((1.8986375231) \right)} + 7}{- \frac{366 (1.8986375231)^{2}}{125} - \sin{\left((1.8986375231) \right)}} = 1.8984642621$