Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{49 x^{3}}{50} - 7$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{49 x_{n}^{3}}{50} + \cos{\left(x_{n} \right)} + 7}{- \frac{147 x_{n}^{2}}{50} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{49 (1.0000000000)^{3}}{50} + \cos{\left((1.0000000000) \right)} + 7}{- \frac{147 (1.0000000000)^{2}}{50} - \sin{\left((1.0000000000) \right)}} = 2.7348545929$ $LaTeX: x_{2} = (2.7348545929) - \frac{- \frac{49 (2.7348545929)^{3}}{50} + \cos{\left((2.7348545929) \right)} + 7}{- \frac{147 (2.7348545929)^{2}}{50} - \sin{\left((2.7348545929) \right)}} = 2.1110270712$ $LaTeX: x_{3} = (2.1110270712) - \frac{- \frac{49 (2.1110270712)^{3}}{50} + \cos{\left((2.1110270712) \right)} + 7}{- \frac{147 (2.1110270712)^{2}}{50} - \sin{\left((2.1110270712) \right)}} = 1.9151866464$ $LaTeX: x_{4} = (1.9151866464) - \frac{- \frac{49 (1.9151866464)^{3}}{50} + \cos{\left((1.9151866464) \right)} + 7}{- \frac{147 (1.9151866464)^{2}}{50} - \sin{\left((1.9151866464) \right)}} = 1.8962599002$ $LaTeX: x_{5} = (1.8962599002) - \frac{- \frac{49 (1.8962599002)^{3}}{50} + \cos{\left((1.8962599002) \right)} + 7}{- \frac{147 (1.8962599002)^{2}}{50} - \sin{\left((1.8962599002) \right)}} = 1.8960905328$