Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{5 x^{3}}{8} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{5 x_{n}^{3}}{8} + \sin{\left(x_{n} \right)} + 4}{- \frac{15 x_{n}^{2}}{8} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{5 (1.0000000000)^{3}}{8} + \sin{\left((1.0000000000) \right)} + 4}{- \frac{15 (1.0000000000)^{2}}{8} + \cos{\left((1.0000000000) \right)}} = 4.1591206034$ $LaTeX: x_{2} = (4.1591206034) - \frac{- \frac{5 (4.1591206034)^{3}}{8} + \sin{\left((4.1591206034) \right)} + 4}{- \frac{15 (4.1591206034)^{2}}{8} + \cos{\left((4.1591206034) \right)}} = 2.8903962244$ $LaTeX: x_{3} = (2.8903962244) - \frac{- \frac{5 (2.8903962244)^{3}}{8} + \sin{\left((2.8903962244) \right)} + 4}{- \frac{15 (2.8903962244)^{2}}{8} + \cos{\left((2.8903962244) \right)}} = 2.2384657382$ $LaTeX: x_{4} = (2.2384657382) - \frac{- \frac{5 (2.2384657382)^{3}}{8} + \sin{\left((2.2384657382) \right)} + 4}{- \frac{15 (2.2384657382)^{2}}{8} + \cos{\left((2.2384657382) \right)}} = 2.0162880030$ $LaTeX: x_{5} = (2.0162880030) - \frac{- \frac{5 (2.0162880030)^{3}}{8} + \sin{\left((2.0162880030) \right)} + 4}{- \frac{15 (2.0162880030)^{2}}{8} + \cos{\left((2.0162880030) \right)}} = 1.9888766831$