Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{293 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{293 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{879 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{293 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{879 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 2.3646868452$ $LaTeX: x_{2} = (2.3646868452) - \frac{- \frac{293 (2.3646868452)^{3}}{500} + \sin{\left((2.3646868452) \right)} + 5}{- \frac{879 (2.3646868452)^{2}}{500} + \cos{\left((2.3646868452) \right)}} = 2.1704947553$ $LaTeX: x_{3} = (2.1704947553) - \frac{- \frac{293 (2.1704947553)^{3}}{500} + \sin{\left((2.1704947553) \right)} + 5}{- \frac{879 (2.1704947553)^{2}}{500} + \cos{\left((2.1704947553) \right)}} = 2.1516710623$ $LaTeX: x_{4} = (2.1516710623) - \frac{- \frac{293 (2.1516710623)^{3}}{500} + \sin{\left((2.1516710623) \right)} + 5}{- \frac{879 (2.1516710623)^{2}}{500} + \cos{\left((2.1516710623) \right)}} = 2.1514989810$ $LaTeX: x_{5} = (2.1514989810) - \frac{- \frac{293 (2.1514989810)^{3}}{500} + \sin{\left((2.1514989810) \right)} + 5}{- \frac{879 (2.1514989810)^{2}}{500} + \cos{\left((2.1514989810) \right)}} = 2.1514989667$