Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{41 x^{3}}{50} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{41 x_{n}^{3}}{50} + \cos{\left(x_{n} \right)} + 7}{- \frac{123 x_{n}^{2}}{50} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{41 (3.0000000000)^{3}}{50} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{123 (3.0000000000)^{2}}{50} - \sin{\left((3.0000000000) \right)}} = 2.2760690445$ $LaTeX: x_{2} = (2.2760690445) - \frac{- \frac{41 (2.2760690445)^{3}}{50} + \cos{\left((2.2760690445) \right)} + 7}{- \frac{123 (2.2760690445)^{2}}{50} - \sin{\left((2.2760690445) \right)}} = 2.0304652557$ $LaTeX: x_{3} = (2.0304652557) - \frac{- \frac{41 (2.0304652557)^{3}}{50} + \cos{\left((2.0304652557) \right)} + 7}{- \frac{123 (2.0304652557)^{2}}{50} - \sin{\left((2.0304652557) \right)}} = 2.0025605755$ $LaTeX: x_{4} = (2.0025605755) - \frac{- \frac{41 (2.0025605755)^{3}}{50} + \cos{\left((2.0025605755) \right)} + 7}{- \frac{123 (2.0025605755)^{2}}{50} - \sin{\left((2.0025605755) \right)}} = 2.0022169407$ $LaTeX: x_{5} = (2.0022169407) - \frac{- \frac{41 (2.0022169407)^{3}}{50} + \cos{\left((2.0022169407) \right)} + 7}{- \frac{123 (2.0022169407)^{2}}{50} - \sin{\left((2.0022169407) \right)}} = 2.0022168890$