Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{149 x^{3}}{200} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{149 x_{n}^{3}}{200} + \cos{\left(x_{n} \right)} + 7}{- \frac{447 x_{n}^{2}}{200} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{149 (3.0000000000)^{3}}{200} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{447 (3.0000000000)^{2}}{200} - \sin{\left((3.0000000000) \right)}} = 2.3036676081$ $LaTeX: x_{2} = (2.3036676081) - \frac{- \frac{149 (2.3036676081)^{3}}{200} + \cos{\left((2.3036676081) \right)} + 7}{- \frac{447 (2.3036676081)^{2}}{200} - \sin{\left((2.3036676081) \right)}} = 2.0833548677$ $LaTeX: x_{3} = (2.0833548677) - \frac{- \frac{149 (2.0833548677)^{3}}{200} + \cos{\left((2.0833548677) \right)} + 7}{- \frac{447 (2.0833548677)^{2}}{200} - \sin{\left((2.0833548677) \right)}} = 2.0618749490$ $LaTeX: x_{4} = (2.0618749490) - \frac{- \frac{149 (2.0618749490)^{3}}{200} + \cos{\left((2.0618749490) \right)} + 7}{- \frac{447 (2.0618749490)^{2}}{200} - \sin{\left((2.0618749490) \right)}} = 2.0616795166$ $LaTeX: x_{5} = (2.0616795166) - \frac{- \frac{149 (2.0616795166)^{3}}{200} + \cos{\left((2.0616795166) \right)} + 7}{- \frac{447 (2.0616795166)^{2}}{200} - \sin{\left((2.0616795166) \right)}} = 2.0616795005$