Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{903 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{903 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 1}{- \frac{2709 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{903 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{2709 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 1.1794979620$ $LaTeX: x_{2} = (1.1794979620) - \frac{- \frac{903 (1.1794979620)^{3}}{1000} + \cos{\left((1.1794979620) \right)} + 1}{- \frac{2709 (1.1794979620)^{2}}{1000} - \sin{\left((1.1794979620) \right)}} = 1.1581104918$ $LaTeX: x_{3} = (1.1581104918) - \frac{- \frac{903 (1.1581104918)^{3}}{1000} + \cos{\left((1.1581104918) \right)} + 1}{- \frac{2709 (1.1581104918)^{2}}{1000} - \sin{\left((1.1581104918) \right)}} = 1.1577716599$ $LaTeX: x_{4} = (1.1577716599) - \frac{- \frac{903 (1.1577716599)^{3}}{1000} + \cos{\left((1.1577716599) \right)} + 1}{- \frac{2709 (1.1577716599)^{2}}{1000} - \sin{\left((1.1577716599) \right)}} = 1.1577715756$ $LaTeX: x_{5} = (1.1577715756) - \frac{- \frac{903 (1.1577715756)^{3}}{1000} + \cos{\left((1.1577715756) \right)} + 1}{- \frac{2709 (1.1577715756)^{2}}{1000} - \sin{\left((1.1577715756) \right)}} = 1.1577715756$