I know this is a rudimentary question but I'm not really sure how to do this. For my homework problem I have to verify some error term of trapazoidal quadrature. I end up with
$$f^{(3)} = -8\sin x\cos x + 6\sin x + 2x\cos x = 0$$
and I need to find the solutions to this. However, I'm not sure where to start. It seems like nothing I know from non-trig zero solving is helping. Can anyone give me some pointers?
Thanks!
This kind of equations, which mix polynomials and trigonometric functions, do not show solutions which can be expressed in terms of elementary functions and I am afraid that only numerical methods could be used.
The particular equation you give $$F(x) = -8\sin (x)\cos( x) + 6\sin( x) + 2x\cos( x)$$ shows an infinite number of zero's (I suppose you looked at the plot). For sure, one of the solutions is trivial $x=0$.
For the range $(0,2\pi)$, there are roots close to $x=2.8$ and $x=5.0$.
To polish the roots, the simplest would be to use Newton method which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}$$ In your case the derivative simplifies to $$F'(x)=-2 x \sin (x)+8 \cos (x)-8 \cos (2 x)$$
Let us apply the method starting with $x_0=2.8$; the successive iterates are $2.75253$, $2.75188$ which is the solution for six significant figures.
Let us apply the method starting with $x_0=5.0$; the successive iterates are $5.03989$, $5.04000$ which is the solution for six significant figures.
I am sure that you can take from here and repeat the process for any root.