Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{17 x^{3}}{40} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{40} + \cos{\left(x_{n} \right)} + 5}{- \frac{51 x_{n}^{2}}{40} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{17 (3.0000000000)^{3}}{40} + \cos{\left((3.0000000000) \right)} + 5}{- \frac{51 (3.0000000000)^{2}}{40} - \sin{\left((3.0000000000) \right)}} = 2.3573592136$ $LaTeX: x_{2} = (2.3573592136) - \frac{- \frac{17 (2.3573592136)^{3}}{40} + \cos{\left((2.3573592136) \right)} + 5}{- \frac{51 (2.3573592136)^{2}}{40} - \sin{\left((2.3573592136) \right)}} = 2.1936572319$ $LaTeX: x_{3} = (2.1936572319) - \frac{- \frac{17 (2.1936572319)^{3}}{40} + \cos{\left((2.1936572319) \right)} + 5}{- \frac{51 (2.1936572319)^{2}}{40} - \sin{\left((2.1936572319) \right)}} = 2.1836203976$ $LaTeX: x_{4} = (2.1836203976) - \frac{- \frac{17 (2.1836203976)^{3}}{40} + \cos{\left((2.1836203976) \right)} + 5}{- \frac{51 (2.1836203976)^{2}}{40} - \sin{\left((2.1836203976) \right)}} = 2.1835838509$ $LaTeX: x_{5} = (2.1835838509) - \frac{- \frac{17 (2.1835838509)^{3}}{40} + \cos{\left((2.1835838509) \right)} + 5}{- \frac{51 (2.1835838509)^{2}}{40} - \sin{\left((2.1835838509) \right)}} = 2.1835838505$