Complex roots in order to apply residue theorem $$\int_{0}^{2\pi}\frac{d\theta}{(4 + 2\sin\theta)^2}$$
$$\sin\theta = \frac{z - z^{-1}}{2i}$$
$$d\theta = \frac{dz}{iz}$$
$$\oint_c\frac{dz}{iz\left(4 +  \frac{z - z^{-1}}{i}\right) ^2}$$
ending up with 
$$\oint_c\frac{dz}{-iz^3 + 8z^2 + 18z - iz^{-1} -8}$$
Now I'm considering De Moivre to get the roots of $z$.
Is that the right way to go?
If so, how do I choose a pole?
 A: $${1\over iz\left(4+{z-{1\over z}\over i}\right)^2}={iz\over(z^2
+4iz-1)^2}={iz\over[\{z-(-2+\sqrt 3)i\}\{(z-(-2-\sqrt3)i\}]^2}$$
So poles are $(-2\pm\sqrt3)i$ and they are of order $2$. As $z=e^{i\theta}\implies|z|=1$, your contour only encircles the pole $(-2+\sqrt3)i$.
A: There is another approach that is a bit more efficient.  We begin with the integral 
$$I(a)=-\frac14 \int_0^{2\pi}\frac{1}{a+\sin(\theta)}\,d\theta \tag 1$$
where $a>1$ (else the integral diverges).
Note that the derivative of $I(a)$ at $a=2$ is given by 
$$I'(2)=\int_0^{2\pi}\frac{1}{(4+2\sin(\theta))^2}\,d\theta \tag 2$$
which is the integral of interest.  Now, when we move to the complex plane, we evaluate $I(a)$ and differentiate.  
Letting $z=e^{i\theta}$ in $(1)$ yields
$$\begin{align}
I(a)&=-\frac14\oint_{|z|=1}\frac{1}{a+\frac{z-z^{-1}}{2i}}\frac{1}{iz}\,dz\\\\
&=-\frac12\oint_{|z|=1}\frac{1}{z^2+2iaz-1}\,dz\\\\
&=-\frac12\oint_{|z|=1}\frac{1}{(z+i(a+\sqrt{a^2-1}))(z+i(a-\sqrt{a^2-1}))}\,dz\\\\
&=-\frac12 (2\pi i)\frac{1}{2i\sqrt{a^2-1}}\\\\
&=\frac{-\pi/2}{\sqrt{a^2-1}} \tag 3
\end{align}$$
where only the pole at $i(a-\sqrt{a^2-1})$ is inside the contour $|z|=1$.
Using $(3)$, we calculate the integral of interest as given by $(2)$ to be 
$$\begin{align}
\int_0^{2\pi}\frac{1}{(4+2\sin(\theta))^2}\,d\theta&=\left.\frac{d}{da}\left(\frac{-\pi/2}{\sqrt{a^2-1}}\right)\right|_{a=2}\\\\
&=\frac{\pi}{3\sqrt 3}
\end{align}$$
