Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{3 x^{3}}{50} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{3 x_{n}^{3}}{50} + \cos{\left(x_{n} \right)} + 3}{- \frac{9 x_{n}^{2}}{50} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{3 (3.0000000000)^{3}}{50} + \cos{\left((3.0000000000) \right)} + 3}{- \frac{9 (3.0000000000)^{2}}{50} - \sin{\left((3.0000000000) \right)}} = 3.2214542459$ $LaTeX: x_{2} = (3.2214542459) - \frac{- \frac{3 (3.2214542459)^{3}}{50} + \cos{\left((3.2214542459) \right)} + 3}{- \frac{9 (3.2214542459)^{2}}{50} - \sin{\left((3.2214542459) \right)}} = 3.2199427216$ $LaTeX: x_{3} = (3.2199427216) - \frac{- \frac{3 (3.2199427216)^{3}}{50} + \cos{\left((3.2199427216) \right)} + 3}{- \frac{9 (3.2199427216)^{2}}{50} - \sin{\left((3.2199427216) \right)}} = 3.2199426177$ $LaTeX: x_{4} = (3.2199426177) - \frac{- \frac{3 (3.2199426177)^{3}}{50} + \cos{\left((3.2199426177) \right)} + 3}{- \frac{9 (3.2199426177)^{2}}{50} - \sin{\left((3.2199426177) \right)}} = 3.2199426177$ $LaTeX: x_{5} = (3.2199426177) - \frac{- \frac{3 (3.2199426177)^{3}}{50} + \cos{\left((3.2199426177) \right)} + 3}{- \frac{9 (3.2199426177)^{2}}{50} - \sin{\left((3.2199426177) \right)}} = 3.2199426177$