Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{27 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{27 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 4}{- \frac{81 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{27 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 4}{- \frac{81 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 4.0960491144$ $LaTeX: x_{2} = (4.0960491144) - \frac{- \frac{27 (4.0960491144)^{3}}{500} + \sin{\left((4.0960491144) \right)} + 4}{- \frac{81 (4.0960491144)^{2}}{500} + \cos{\left((4.0960491144) \right)}} = 3.9361642064$ $LaTeX: x_{3} = (3.9361642064) - \frac{- \frac{27 (3.9361642064)^{3}}{500} + \sin{\left((3.9361642064) \right)} + 4}{- \frac{81 (3.9361642064)^{2}}{500} + \cos{\left((3.9361642064) \right)}} = 3.9340686933$ $LaTeX: x_{4} = (3.9340686933) - \frac{- \frac{27 (3.9340686933)^{3}}{500} + \sin{\left((3.9340686933) \right)} + 4}{- \frac{81 (3.9340686933)^{2}}{500} + \cos{\left((3.9340686933) \right)}} = 3.9340683088$ $LaTeX: x_{5} = (3.9340683088) - \frac{- \frac{27 (3.9340683088)^{3}}{500} + \sin{\left((3.9340683088) \right)} + 4}{- \frac{81 (3.9340683088)^{2}}{500} + \cos{\left((3.9340683088) \right)}} = 3.9340683088$