Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{337 x^{3}}{500} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{337 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 1}{- \frac{1011 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{337 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{1011 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 1.7879279218$ $LaTeX: x_{2} = (1.7879279218) - \frac{- \frac{337 (1.7879279218)^{3}}{500} + \sin{\left((1.7879279218) \right)} + 1}{- \frac{1011 (1.7879279218)^{2}}{500} + \cos{\left((1.7879279218) \right)}} = 1.5070993520$ $LaTeX: x_{3} = (1.5070993520) - \frac{- \frac{337 (1.5070993520)^{3}}{500} + \sin{\left((1.5070993520) \right)} + 1}{- \frac{1011 (1.5070993520)^{2}}{500} + \cos{\left((1.5070993520) \right)}} = 1.4388218713$ $LaTeX: x_{4} = (1.4388218713) - \frac{- \frac{337 (1.4388218713)^{3}}{500} + \sin{\left((1.4388218713) \right)} + 1}{- \frac{1011 (1.4388218713)^{2}}{500} + \cos{\left((1.4388218713) \right)}} = 1.4347981703$ $LaTeX: x_{5} = (1.4347981703) - \frac{- \frac{337 (1.4347981703)^{3}}{500} + \sin{\left((1.4347981703) \right)} + 1}{- \frac{1011 (1.4347981703)^{2}}{500} + \cos{\left((1.4347981703) \right)}} = 1.4347844923$