Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{963 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{963 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{2889 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{963 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{2889 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 1.3740247146$ $LaTeX: x_{2} = (1.3740247146) - \frac{- \frac{963 (1.3740247146)^{3}}{1000} + \sin{\left((1.3740247146) \right)} + 1}{- \frac{2889 (1.3740247146)^{2}}{1000} + \cos{\left((1.3740247146) \right)}} = 1.2756370398$ $LaTeX: x_{3} = (1.2756370398) - \frac{- \frac{963 (1.2756370398)^{3}}{1000} + \sin{\left((1.2756370398) \right)} + 1}{- \frac{2889 (1.2756370398)^{2}}{1000} + \cos{\left((1.2756370398) \right)}} = 1.2660637341$ $LaTeX: x_{4} = (1.2660637341) - \frac{- \frac{963 (1.2660637341)^{3}}{1000} + \sin{\left((1.2660637341) \right)} + 1}{- \frac{2889 (1.2660637341)^{2}}{1000} + \cos{\left((1.2660637341) \right)}} = 1.2659758270$ $LaTeX: x_{5} = (1.2659758270) - \frac{- \frac{963 (1.2659758270)^{3}}{1000} + \sin{\left((1.2659758270) \right)} + 1}{- \frac{2889 (1.2659758270)^{2}}{1000} + \cos{\left((1.2659758270) \right)}} = 1.2659758196$