Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{213 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{213 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 7}{- \frac{639 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{213 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{639 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.6509059709$ $LaTeX: x_{2} = (2.6509059709) - \frac{- \frac{213 (2.6509059709)^{3}}{500} + \sin{\left((2.6509059709) \right)} + 7}{- \frac{639 (2.6509059709)^{2}}{500} + \cos{\left((2.6509059709) \right)}} = 2.6037999330$ $LaTeX: x_{3} = (2.6037999330) - \frac{- \frac{213 (2.6037999330)^{3}}{500} + \sin{\left((2.6037999330) \right)} + 7}{- \frac{639 (2.6037999330)^{2}}{500} + \cos{\left((2.6037999330) \right)}} = 2.6029587254$ $LaTeX: x_{4} = (2.6029587254) - \frac{- \frac{213 (2.6029587254)^{3}}{500} + \sin{\left((2.6029587254) \right)} + 7}{- \frac{639 (2.6029587254)^{2}}{500} + \cos{\left((2.6029587254) \right)}} = 2.6029584590$ $LaTeX: x_{5} = (2.6029584590) - \frac{- \frac{213 (2.6029584590)^{3}}{500} + \sin{\left((2.6029584590) \right)} + 7}{- \frac{639 (2.6029584590)^{2}}{500} + \cos{\left((2.6029584590) \right)}} = 2.6029584590$