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

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- x_{n}^{3} + \sin{\left(x_{n} \right)} + 7}{- 3 x_{n}^{2} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- (1.0000000000)^{3} + \sin{\left((1.0000000000) \right)} + 7}{- 3 (1.0000000000)^{2} + \cos{\left((1.0000000000) \right)}} = 3.7814275718$ $LaTeX: x_{2} = (3.7814275718) - \frac{- (3.7814275718)^{3} + \sin{\left((3.7814275718) \right)} + 7}{- 3 (3.7814275718)^{2} + \cos{\left((3.7814275718) \right)}} = 2.6906112402$ $LaTeX: x_{3} = (2.6906112402) - \frac{- (2.6906112402)^{3} + \sin{\left((2.6906112402) \right)} + 7}{- 3 (2.6906112402)^{2} + \cos{\left((2.6906112402) \right)}} = 2.1581842899$ $LaTeX: x_{4} = (2.1581842899) - \frac{- (2.1581842899)^{3} + \sin{\left((2.1581842899) \right)} + 7}{- 3 (2.1581842899)^{2} + \cos{\left((2.1581842899) \right)}} = 2.0053763604$ $LaTeX: x_{5} = (2.0053763604) - \frac{- (2.0053763604)^{3} + \sin{\left((2.0053763604) \right)} + 7}{- 3 (2.0053763604)^{2} + \cos{\left((2.0053763604) \right)}} = 1.9927504105$