Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{143 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{143 x_{n}^{3}}{250} + \sin{\left(x_{n} \right)} + 6}{- \frac{429 x_{n}^{2}}{250} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{143 (3.0000000000)^{3}}{250} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{429 (3.0000000000)^{2}}{250} + \cos{\left((3.0000000000) \right)}} = 2.4339245321$ $LaTeX: x_{2} = (2.4339245321) - \frac{- \frac{143 (2.4339245321)^{3}}{250} + \sin{\left((2.4339245321) \right)} + 6}{- \frac{429 (2.4339245321)^{2}}{250} + \cos{\left((2.4339245321) \right)}} = 2.2877206836$ $LaTeX: x_{3} = (2.2877206836) - \frac{- \frac{143 (2.2877206836)^{3}}{250} + \sin{\left((2.2877206836) \right)} + 6}{- \frac{429 (2.2877206836)^{2}}{250} + \cos{\left((2.2877206836) \right)}} = 2.2778825184$ $LaTeX: x_{4} = (2.2778825184) - \frac{- \frac{143 (2.2778825184)^{3}}{250} + \sin{\left((2.2778825184) \right)} + 6}{- \frac{429 (2.2778825184)^{2}}{250} + \cos{\left((2.2778825184) \right)}} = 2.2778389731$ $LaTeX: x_{5} = (2.2778389731) - \frac{- \frac{143 (2.2778389731)^{3}}{250} + \sin{\left((2.2778389731) \right)} + 6}{- \frac{429 (2.2778389731)^{2}}{250} + \cos{\left((2.2778389731) \right)}} = 2.2778389723$