Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{213 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{213 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 6}{- \frac{639 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{213 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{639 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.2972040816$ $LaTeX: x_{2} = (2.2972040816) - \frac{- \frac{213 (2.2972040816)^{3}}{250} + \sin{\left((2.2972040816) \right)} + 6}{- \frac{639 (2.2972040816)^{2}}{250} + \cos{\left((2.2972040816) \right)}} = 2.0441787163$ $LaTeX: x_{3} = (2.0441787163) - \frac{- \frac{213 (2.0441787163)^{3}}{250} + \sin{\left((2.0441787163) \right)} + 6}{- \frac{639 (2.0441787163)^{2}}{250} + \cos{\left((2.0441787163) \right)}} = 2.0093652453$ $LaTeX: x_{4} = (2.0093652453) - \frac{- \frac{213 (2.0093652453)^{3}}{250} + \sin{\left((2.0093652453) \right)} + 6}{- \frac{639 (2.0093652453)^{2}}{250} + \cos{\left((2.0093652453) \right)}} = 2.0087287370$ $LaTeX: x_{5} = (2.0087287370) - \frac{- \frac{213 (2.0087287370)^{3}}{250} + \sin{\left((2.0087287370) \right)} + 6}{- \frac{639 (2.0087287370)^{2}}{250} + \cos{\left((2.0087287370) \right)}} = 2.0087285261$