I endeavored to find $\displaystyle \int _0^{2\pi }\frac{d\theta }{3 + \cos\theta +2\sin\theta }$
I wrote and simplified: $\displaystyle f\left(z\right)=\frac{1}{z^2\left(1+\frac{i}{2}\right)+3zi+\frac{i}{2}-1}$ ,
which has a pole within the unit circle: $\displaystyle -\frac15 - \frac{2i}5$. The other one is $-2i-1$.
Finally the residue theorem: $\displaystyle 2i\pi \lim _{z\to \frac{−1}{5}-\frac{2i}{5}}\frac{1}{\left(1+\frac{i}{2}\right)\left(z+2i+1\right)}$
Leading me to $\displaystyle \int _0^{2\pi }\frac{d\theta }{3 + \cos\theta + 2\sin\theta }=5\pi $
Did I stumble along the way?
Your method is fine, but you clearly have an arithmetic error near the end. As you've found, your integral is $$\frac{2i\pi}{(1+\frac{i}{2})(2i+1)\frac{4}{5}}=\frac{2i\pi}{2i}=\pi.$$