Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{803 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{803 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 1}{- \frac{2409 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{803 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{2409 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 1.5557190915$ $LaTeX: x_{2} = (1.5557190915) - \frac{- \frac{803 (1.5557190915)^{3}}{1000} + \sin{\left((1.5557190915) \right)} + 1}{- \frac{2409 (1.5557190915)^{2}}{1000} + \cos{\left((1.5557190915) \right)}} = 1.3797004026$ $LaTeX: x_{3} = (1.3797004026) - \frac{- \frac{803 (1.3797004026)^{3}}{1000} + \sin{\left((1.3797004026) \right)} + 1}{- \frac{2409 (1.3797004026)^{2}}{1000} + \cos{\left((1.3797004026) \right)}} = 1.3507701132$ $LaTeX: x_{4} = (1.3507701132) - \frac{- \frac{803 (1.3507701132)^{3}}{1000} + \sin{\left((1.3507701132) \right)} + 1}{- \frac{2409 (1.3507701132)^{2}}{1000} + \cos{\left((1.3507701132) \right)}} = 1.3500106416$ $LaTeX: x_{5} = (1.3500106416) - \frac{- \frac{803 (1.3500106416)^{3}}{1000} + \sin{\left((1.3500106416) \right)} + 1}{- \frac{2409 (1.3500106416)^{2}}{1000} + \cos{\left((1.3500106416) \right)}} = 1.3500101243$