1. The problem is to solve the equation $2\cot(2a) = 3$ for $a$. 2. Start by isolating $\cot(2a)$: $$2\cot(2a) = 3 \implies \cot(2a) = \frac{3}{2}$$ 3. Recall that $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$, so we're looking for angles $2a$ where the cotangent is $\frac{3}{2}$. 4. Taking the arccotangent (inverse cotangent), we get: $$2a = \arccot\left(\frac{3}{2}\right) + k\pi, \quad k \in \mathbb{Z}$$ Because cotangent has period $\pi$. 5. We can express $\arccot(x)$ as $\arctan\left(\frac{1}{x}\right)$, hence: $$2a = \arctan\left(\frac{2}{3}\right) + k\pi$$ 6. Divide both sides by 2 to solve for $a$: $$a = \frac{1}{2} \arctan\left(\frac{2}{3}\right) + \frac{k\pi}{2}$$ 7. This is the general solution. For any integer $k$, $a$ takes these values. **Final answer:** $$a = \frac{1}{2} \arctan\left(\frac{2}{3}\right) + \frac{k\pi}{2}, \quad k \in \mathbb{Z}$$