Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{21 x^{3}}{100} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{21 x_{n}^{3}}{100} + \cos{\left(x_{n} \right)} + 2}{- \frac{63 x_{n}^{2}}{100} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{21 (1.0000000000)^{3}}{100} + \cos{\left((1.0000000000) \right)} + 2}{- \frac{63 (1.0000000000)^{2}}{100} - \sin{\left((1.0000000000) \right)}} = 2.5836549480$ $LaTeX: x_{2} = (2.5836549480) - \frac{- \frac{21 (2.5836549480)^{3}}{100} + \cos{\left((2.5836549480) \right)} + 2}{- \frac{63 (2.5836549480)^{2}}{100} - \sin{\left((2.5836549480) \right)}} = 2.0619636389$ $LaTeX: x_{3} = (2.0619636389) - \frac{- \frac{21 (2.0619636389)^{3}}{100} + \cos{\left((2.0619636389) \right)} + 2}{- \frac{63 (2.0619636389)^{2}}{100} - \sin{\left((2.0619636389) \right)}} = 1.9741375532$ $LaTeX: x_{4} = (1.9741375532) - \frac{- \frac{21 (1.9741375532)^{3}}{100} + \cos{\left((1.9741375532) \right)} + 2}{- \frac{63 (1.9741375532)^{2}}{100} - \sin{\left((1.9741375532) \right)}} = 1.9717199525$ $LaTeX: x_{5} = (1.9717199525) - \frac{- \frac{21 (1.9717199525)^{3}}{100} + \cos{\left((1.9717199525) \right)} + 2}{- \frac{63 (1.9717199525)^{2}}{100} - \sin{\left((1.9717199525) \right)}} = 1.9717181360$