Find zeroes of trigonometric polynomial 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!
 A: 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.
