Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{667 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{667 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{2001 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{667 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{2001 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4806103537$ $LaTeX: x_{2} = (2.4806103537) - \frac{- \frac{667 (2.4806103537)^{3}}{1000} + \sin{\left((2.4806103537) \right)} + 8}{- \frac{2001 (2.4806103537)^{2}}{1000} + \cos{\left((2.4806103537) \right)}} = 2.3609859548$ $LaTeX: x_{3} = (2.3609859548) - \frac{- \frac{667 (2.3609859548)^{3}}{1000} + \sin{\left((2.3609859548) \right)} + 8}{- \frac{2001 (2.3609859548)^{2}}{1000} + \cos{\left((2.3609859548) \right)}} = 2.3547066729$ $LaTeX: x_{4} = (2.3547066729) - \frac{- \frac{667 (2.3547066729)^{3}}{1000} + \sin{\left((2.3547066729) \right)} + 8}{- \frac{2001 (2.3547066729)^{2}}{1000} + \cos{\left((2.3547066729) \right)}} = 2.3546897237$ $LaTeX: x_{5} = (2.3546897237) - \frac{- \frac{667 (2.3546897237)^{3}}{1000} + \sin{\left((2.3546897237) \right)} + 8}{- \frac{2001 (2.3546897237)^{2}}{1000} + \cos{\left((2.3546897237) \right)}} = 2.3546897236$