Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{77 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{77 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 7}{- \frac{231 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{77 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{231 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 2.4614184523$ $LaTeX: x_{2} = (2.4614184523) - \frac{- \frac{77 (2.4614184523)^{3}}{125} + \sin{\left((2.4614184523) \right)} + 7}{- \frac{231 (2.4614184523)^{2}}{125} + \cos{\left((2.4614184523) \right)}} = 2.3313589344$ $LaTeX: x_{3} = (2.3313589344) - \frac{- \frac{77 (2.3313589344)^{3}}{125} + \sin{\left((2.3313589344) \right)} + 7}{- \frac{231 (2.3313589344)^{2}}{125} + \cos{\left((2.3313589344) \right)}} = 2.3237953420$ $LaTeX: x_{4} = (2.3237953420) - \frac{- \frac{77 (2.3237953420)^{3}}{125} + \sin{\left((2.3237953420) \right)} + 7}{- \frac{231 (2.3237953420)^{2}}{125} + \cos{\left((2.3237953420) \right)}} = 2.3237703045$ $LaTeX: x_{5} = (2.3237703045) - \frac{- \frac{77 (2.3237703045)^{3}}{125} + \sin{\left((2.3237703045) \right)} + 7}{- \frac{231 (2.3237703045)^{2}}{125} + \cos{\left((2.3237703045) \right)}} = 2.3237703042$