Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{46 x^{3}}{125} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{46 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 5}{- \frac{138 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{46 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{138 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 2.5611492509$ $LaTeX: x_{2} = (2.5611492509) - \frac{- \frac{46 (2.5611492509)^{3}}{125} + \sin{\left((2.5611492509) \right)} + 5}{- \frac{138 (2.5611492509)^{2}}{125} + \cos{\left((2.5611492509) \right)}} = 2.4826709159$ $LaTeX: x_{3} = (2.4826709159) - \frac{- \frac{46 (2.4826709159)^{3}}{125} + \sin{\left((2.4826709159) \right)} + 5}{- \frac{138 (2.4826709159)^{2}}{125} + \cos{\left((2.4826709159) \right)}} = 2.4801704909$ $LaTeX: x_{4} = (2.4801704909) - \frac{- \frac{46 (2.4801704909)^{3}}{125} + \sin{\left((2.4801704909) \right)} + 5}{- \frac{138 (2.4801704909)^{2}}{125} + \cos{\left((2.4801704909) \right)}} = 2.4801679782$ $LaTeX: x_{5} = (2.4801679782) - \frac{- \frac{46 (2.4801679782)^{3}}{125} + \sin{\left((2.4801679782) \right)} + 5}{- \frac{138 (2.4801679782)^{2}}{125} + \cos{\left((2.4801679782) \right)}} = 2.4801679782$