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Solve the equation

2 + \log_{2} (\cos θ) = \frac{1}{2} \log_{2} (3) - \log_{2} (\sin θ), 0 \leq θ \leq 2 π

Recall that

\sin (2 θ) = 2 \sin θ \cos θ

Once you have your solutions, you need to select only the ones where both $\cos θ$ and $\sin θ$ are positive, so that $\log (\cos θ)$ and $\log (\sin θ)$ are defined.

The product of the solutions can be expressed as $\frac{π^{n}}{m}$ for $n, m \in N$ .

Give the value of

n m