Alright — let’s go step by step.  

We are to compute  

\[
I = \int_{0}^{2\pi} \frac{d\theta}{3 + 2\cos\theta}.
\]

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### 1. Convert to a complex contour integral
Let \( z = e^{i\theta} \), \( dz = i e^{i\theta} d\theta = iz \, d\theta \), so \( d\theta = \frac{dz}{iz} \).

Also,  

\[
\cos\theta = \frac{z + z^{-1}}{2}.
\]

The integral becomes:

\[
I = \oint_{|z|=1} \frac{1}{3 + 2 \cdot \frac{z+z^{-1}}{2}} \cdot \frac{dz}{iz}.
\]

Simplify denominator:  
\( 3 + 2\cdot \frac{z+z^{-1}}{2} = 3 + z + z^{-1} \).

So:

\[
I = \oint_{|z|=1} \frac{1}{3 + z + \frac{1}{z}} \cdot \frac{dz}{iz}
= \oint_{|z|=1} \frac{1}{\frac{3z + z^2 + 1}{z}} \cdot \frac{dz}{iz}.
\]

\[
= \oint_{|z|=1} \frac{z}{z^2 + 3z + 1} \cdot \frac{dz}{iz}
= \frac{1}{i} \oint_{|z|=1} \frac{1}{z^2 + 3z + 1} \, dz.
\]

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### 2. Find poles (zeros of denominator)
Solve \( z^2 + 3z + 1 = 0 \).

\[
z = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm