Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{183 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{183 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 6}{- \frac{549 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{183 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{549 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.3436019602$ $LaTeX: x_{2} = (2.3436019602) - \frac{- \frac{183 (2.3436019602)^{3}}{250} + \sin{\left((2.3436019602) \right)} + 6}{- \frac{549 (2.3436019602)^{2}}{250} + \cos{\left((2.3436019602) \right)}} = 2.1314897626$ $LaTeX: x_{3} = (2.1314897626) - \frac{- \frac{183 (2.1314897626)^{3}}{250} + \sin{\left((2.1314897626) \right)} + 6}{- \frac{549 (2.1314897626)^{2}}{250} + \cos{\left((2.1314897626) \right)}} = 2.1084880323$ $LaTeX: x_{4} = (2.1084880323) - \frac{- \frac{183 (2.1084880323)^{3}}{250} + \sin{\left((2.1084880323) \right)} + 6}{- \frac{549 (2.1084880323)^{2}}{250} + \cos{\left((2.1084880323) \right)}} = 2.1082259694$ $LaTeX: x_{5} = (2.1082259694) - \frac{- \frac{183 (2.1082259694)^{3}}{250} + \sin{\left((2.1082259694) \right)} + 6}{- \frac{549 (2.1082259694)^{2}}{250} + \cos{\left((2.1082259694) \right)}} = 2.1082259355$