Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{93 x^{3}}{100} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{93 x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 6}{- \frac{279 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{93 (1.0000000000)^{3}}{100} + \sin{\left((1.0000000000) \right)} + 6}{- \frac{279 (1.0000000000)^{2}}{100} + \cos{\left((1.0000000000) \right)}} = 3.6276734871$ $LaTeX: x_{2} = (3.6276734871) - \frac{- \frac{93 (3.6276734871)^{3}}{100} + \sin{\left((3.6276734871) \right)} + 6}{- \frac{279 (3.6276734871)^{2}}{100} + \cos{\left((3.6276734871) \right)}} = 2.5940311433$ $LaTeX: x_{3} = (2.5940311433) - \frac{- \frac{93 (2.5940311433)^{3}}{100} + \sin{\left((2.5940311433) \right)} + 6}{- \frac{279 (2.5940311433)^{2}}{100} + \cos{\left((2.5940311433) \right)}} = 2.0991816947$ $LaTeX: x_{4} = (2.0991816947) - \frac{- \frac{93 (2.0991816947)^{3}}{100} + \sin{\left((2.0991816947) \right)} + 6}{- \frac{279 (2.0991816947)^{2}}{100} + \cos{\left((2.0991816947) \right)}} = 1.9633025034$ $LaTeX: x_{5} = (1.9633025034) - \frac{- \frac{93 (1.9633025034)^{3}}{100} + \sin{\left((1.9633025034) \right)} + 6}{- \frac{279 (1.9633025034)^{2}}{100} + \cos{\left((1.9633025034) \right)}} = 1.9530686754$