Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{69 x^{3}}{125} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{69 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 3}{- \frac{207 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{69 (1.0000000000)^{3}}{125} + \sin{\left((1.0000000000) \right)} + 3}{- \frac{207 (1.0000000000)^{2}}{125} + \cos{\left((1.0000000000) \right)}} = 3.9483533058$ $LaTeX: x_{2} = (3.9483533058) - \frac{- \frac{69 (3.9483533058)^{3}}{125} + \sin{\left((3.9483533058) \right)} + 3}{- \frac{207 (3.9483533058)^{2}}{125} + \cos{\left((3.9483533058) \right)}} = 2.7525195749$ $LaTeX: x_{3} = (2.7525195749) - \frac{- \frac{69 (2.7525195749)^{3}}{125} + \sin{\left((2.7525195749) \right)} + 3}{- \frac{207 (2.7525195749)^{2}}{125} + \cos{\left((2.7525195749) \right)}} = 2.1488752770$ $LaTeX: x_{4} = (2.1488752770) - \frac{- \frac{69 (2.1488752770)^{3}}{125} + \sin{\left((2.1488752770) \right)} + 3}{- \frac{207 (2.1488752770)^{2}}{125} + \cos{\left((2.1488752770) \right)}} = 1.9487275731$ $LaTeX: x_{5} = (1.9487275731) - \frac{- \frac{69 (1.9487275731)^{3}}{125} + \sin{\left((1.9487275731) \right)} + 3}{- \frac{207 (1.9487275731)^{2}}{125} + \cos{\left((1.9487275731) \right)}} = 1.9253601497$