Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{37 x^{3}}{40} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{37 x_{n}^{3}}{40} + \sin{\left(x_{n} \right)} + 7}{- \frac{111 x_{n}^{2}}{40} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{37 (3.0000000000)^{3}}{40} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{111 (3.0000000000)^{2}}{40} + \cos{\left((3.0000000000) \right)}} = 2.3131567439$ $LaTeX: x_{2} = (2.3131567439) - \frac{- \frac{37 (2.3131567439)^{3}}{40} + \sin{\left((2.3131567439) \right)} + 7}{- \frac{111 (2.3131567439)^{2}}{40} + \cos{\left((2.3131567439) \right)}} = 2.0740562778$ $LaTeX: x_{3} = (2.0740562778) - \frac{- \frac{37 (2.0740562778)^{3}}{40} + \sin{\left((2.0740562778) \right)} + 7}{- \frac{111 (2.0740562778)^{2}}{40} + \cos{\left((2.0740562778) \right)}} = 2.0437150739$ $LaTeX: x_{4} = (2.0437150739) - \frac{- \frac{37 (2.0437150739)^{3}}{40} + \sin{\left((2.0437150739) \right)} + 7}{- \frac{111 (2.0437150739)^{2}}{40} + \cos{\left((2.0437150739) \right)}} = 2.0432437106$ $LaTeX: x_{5} = (2.0432437106) - \frac{- \frac{37 (2.0432437106)^{3}}{40} + \sin{\left((2.0432437106) \right)} + 7}{- \frac{111 (2.0432437106)^{2}}{40} + \cos{\left((2.0432437106) \right)}} = 2.0432435977$